[Guest]

I recently came across this very cool weighing problem. It's not your usual balance scale problem and it's quite hard, but doable.

There are 13 separate piles of coins. Each pile has 4 identical coins in it. 12 of the piles are all identical to each other, containing the genuine coins. A genuine coin weighs X grams, where X is an integer (important). The 1 pile of counterfeit coins look and feel identical to the genuine coins. But, they weigh X-delta, where delta can be any real number (important), such that 5>=delta>=-5. For example, if X=8, then the counterfeit coins can weigh anywhere from 3 to 13.

You have a digital readout scale that will display a single number for each weighing, accurate to any real number. Your task is to determine X, delta and the pile that contains the counterfeit coins by using the scale only 2 times.

You can return coins to their original piles after each weighing, and you can decide what your next weighing will be based on the result from the first weighing.

Edit on behalf of OP:

Of course if delta=0 the counterfeit pile cannot be identified by weighing.

Edited May 30, 2010 by bonanova
OP requested the delta=0 caveat

[Guest]

I recently came across this very cool weighing problem. It's not your usual balance scale problem and it's quite hard, but doable.

There are 13 separate piles of coins. Each pile has 4 identical coins in it. 12 of the piles are all identical to each other, containing the genuine coins. A genuine coin weighs X grams, where X is an integer (important). The 1 pile of counterfeit coins look and feel identical to the genuine coins. But, they weigh X-delta, where delta can be any real number (important), such that 5>=delta>=-5. For example, if X=8, then the counterfeit coins can weigh anywhere from 3 to 13.

You have a digital readout scale that will display a single number for each weighing, accurate to any real number. Your task is to determine X, delta and the pile that contains the counterfeit coins by using the scale only 2 times.

You can return coins to their original piles after each weighing, and you can decide what your next weighing will be based on the result from the first weighing.

I could find out the way to estimate the actual weight of genuine coin and whether the fake coin is 5 gms less or 5 gms more than the genuine coin. That can be done in a single weighing. Take 1 coin each from each pile and weigh all 13 coins. If the total weight divided by 13 happens to contain decimal (0.615), then the fake coin is 5 gms less and if it contains (0.384), then the fake coin is 5 gms more. The weight of genuine coin is obtained by rounding off to higher integer in case of decimal (0.615) OR rounding off to lower integer (if the decimal is 0.384).

Having done this I am stuck now to identify the pile which contains fake coins. The limitation is that each pile contains max of 4 coins. Any body help ??

[Guest]

Can Δ be 0?

[bushindo]

I could find out the way to estimate the actual weight of genuine coin and whether the fake coin is 5 gms less or 5 gms more than the genuine coin. That can be done in a single weighing. Take 1 coin each from each pile and weigh all 13 coins. If the total weight divided by 13 happens to contain decimal (0.615), then the fake coin is 5 gms less and if it contains (0.384), then the fake coin is 5 gms more. The weight of genuine coin is obtained by rounding off to higher integer in case of decimal (0.615) OR rounding off to lower integer (if the decimal is 0.384).

Having done this I am stuck now to identify the pile which contains fake coins. The limitation is that each pile contains max of 4 coins. Any body help ??

Interesting problem. Here's my attempt

Label the piles from 1-13. We will only take coins from pile 1-12. For the first and second weighing, draw the following number of coin from each numbered pile and weigh those drawn coins. Note that the ratio of the first and second draw for every pile is a unique ratio. There are 12 such unique ratios.

     1st  2nd         ratio         

P 1  1     1      1.0000000

P 2  1     2      2.0000000

P 3  1     3      3.0000000

P 4  1     4      4.0000000

P 5  2     1      0.5000000

P 6  2     3      1.5000000

P 7  3     1      0.3333333

P 8  3     2      0.6666667

P 9  3     0      0.0000000

P 10 3     4      1.3333333

P 11 4     1      0.2500000

P 12 4     3      0.7500000

Now, for the first weighing, we will be putting 30 coins onto the register. If we get an integer as a total weight, then pile 13 is the odd pile and it is trivial to figure out delta. If the first weighting returns a real number (call it K₁), do the 2nd weighting as directed. There will be 25 coins for the second weighting and we'll get a second number (K₂). So, we have the following equation

30X - a * delta = K₁

25X - b * delta = K₂,

where a and b are the first and second draw number for some pile in the matrix above. X is either floor( K₁/30 ) or ceil( K₁/30 ). We can compute the ratio b/a by substituting these two possible values for X in

b/a = (K₂ - 25X )/ (K₁ - 30X ).

If we substitute the correct X, then the ratio b/a will be one of those in the matrix above, which will tell us which pile the odd coins are in. The rest of the parameters can then be computed.

Edited May 28, 2010 by bushindo

[Guest]

Here goes,

Call the piles 1,2,3,4,5,6,7,8,9,0,A,B,C.

X is the normal coin weight and R is delta.

Put the following on the scale:

1 coin from each of 1,2 and 3

2 coins from each of 4,5 and 6

3 coins from each of 7,8,9 and 0

4 coins from each of A,B and C

Divide the total weight by 33 and leave the remainder. Call the quotient x and the remainder r.

If r=1, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 1, then X=x, R=1 and the fakes are in pile 1.

If the difference is 2, then X=x, R=1 and the fakes are in pile 2.

If the difference is 3, then X=x, R=1 and the fakes are in pile 3.

If r=2, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 4, 2 from 5 and 3 from 6 on the scale. Subtract 15x from the result.

If the difference is 1, the X=x, R=1 and the fakes are in pile 4.

If the difference is 2, the X=x, R=1 and the fakes are in pile 5.

If the difference is 3, the X=x, R=1 and the fakes are in pile 6.

If the difference is 4, the X=x, R=2 and the fakes are in pile 1.

If the difference is 6, the X=x, R=2 and the fakes are in pile 2.

If the difference is 8, the X=x, R=2 and the fakes are in pile 3.

If r=3, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 7, 2 from 8, 3 from 9, 4 from 10. Subtract 19x from the result.

If the difference is 1, the X=x, R=1 and the fakes are in pile 7.

If the difference is 2, the X=x, R=1 and the fakes are in pile 8.

If the difference is 3, the X=x, R=1 and the fakes are in pile 9.

If the difference is 4, the X=x, R=1 and the fakes are in pile 0.

If the difference is 6, the X=x, R=3 and the fakes are in pile 1.

If the difference is 9, the X=x, R=3 and the fakes are in pile 2.

If the difference is 12, the X=x, R=3 and the fakes are in pile 3.

If r=4, then put the following on the scale: 3 from 1, 4 from 2, 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 22x from the result.

If the difference is 1, the X=x, R=1 and the fakes are in pile A.

If the difference is 2, the X=x, R=1 and the fakes are in pile B.

If the difference is 3, the X=x, R=1 and the fakes are in pile C.

If the difference is 4, the X=x, R=2 and the fakes are in pile 4.

If the difference is 6, the X=x, R=2 and the fakes are in pile 5.

If the difference is 8, the X=x, R=2 and the fakes are in pile 6.

If the difference is 12, the X=x, R=3 and the fakes are in pile 1.

If the difference is 16, the X=x, R=3 and the fakes are in pile 2.

If the difference is 0, the X=x, R=3 and the fakes are in pile 3.

If r=5, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 5, the X=x, R=5 and the fakes are in pile 1.

If the difference is 10, the X=x, R=5 and the fakes are in pile 2.

If the difference is 15, the X=x, R=5 and the fakes are in pile 3.

If r=6, then put the following on the scale: 1 from 4, 3 from 5, 4 from 6, 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 18x from the result.

If the difference is 2, the X=x, R=2 and the fakes are in pile 7.

If the difference is 3, the X=x, R=3 and the fakes are in pile 4.

If the difference is 4, the X=x, R=2 and the fakes are in pile 8.

If the difference is 6, the X=x, R=2 and the fakes are in pile 9.

If the difference is 8, the X=x, R=2 and the fakes are in pile 0.

If the difference is 9, the X=x, R=3 and the fakes are in pile 5.

If the difference is 12, the X=x, R=3 and the fakes are in pile 6.

r will not equal 7.

If r=8, then put the following on the scale: 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 15x from the result.

If the difference is 2, the X=x, R=2 and the fakes are in pile A.

If the difference is 4, the X=x, R=2 and the fakes are in pile B.

If the difference is 6, the X=x, R=2 and the fakes are in pile C.

If the difference is 8, the X=x, R=4 and the fakes are in pile 4.

If the difference is 12, the X=x, R=4 and the fakes are in pile 5.

If the difference is 16, the X=x, R=4 and the fakes are in pile 6.

If r=9, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 3, the X=x, R=3 and the fakes are in pile 7.

If the difference is 6, the X=x, R=3 and the fakes are in pile 8.

If the difference is 9, the X=x, R=3 and the fakes are in pile 9.

If the difference is 12, the X=x, R=3 and the fakes are in pile 0.

If r=10, then put the following on the scale: 1 from 4, 2 from 5 and 3 from 6. Subtract 6x from the result.

If the difference is 5, the X=x, R=5 and the fakes are in pile 4.

If the difference is 10, the X=x, R=5 and the fakes are in pile 5.

If the difference is 15, the X=x, R=5 and the fakes are in pile 6.

r will not equal 11.

If r=12, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9, 4 from 0, 1 from A, 2 from B and 3 from C. Subtract 16x from the result.

If the difference is 3, the X=x, R=3 and the fakes are in pile A.

If the difference is 4, the X=x, R=4 and the fakes are in pile 7.

If the difference is 6, the X=x, R=3 and the fakes are in pile B.

If the difference is 8, the X=x, R=4 and the fakes are in pile 8.

If the difference is 9, the X=x, R=3 and the fakes are in pile C.

If the difference is 12, the X=x, R=4 and the fakes are in pile 9.

If the difference is 16, the X=x, R=4 and the fakes are in pile 0.

If the difference is , the X=x, R= and the fakes are in pile .

If r=13, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is -5, the X=x+1, R=-5 and the fakes are in pile A.

If the difference is -10, the X=x+1, R=-5 and the fakes are in pile B.

If the difference is -15, the X=x+1, R=-5 and the fakes are in pile C.

r will not equal 14.

If r=15, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 5, the X=x, R=5 and the fakes are in pile 7.

If the difference is 10, the X=x, R=5 and the fakes are in pile 8.

If the difference is 15, the X=x, R=5 and the fakes are in pile 9.

If the difference is 20, the X=x, R=5 and the fakes are in pile 0.

If r=16, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is 4, the X=x, R=4 and the fakes are in pile A.

If the difference is 8, the X=x, R=4 and the fakes are in pile B.

If the difference is 12, the X=x, R=4 and the fakes are in pile C.

If the difference is , the X=x, R= and the fakes are in pile .

If r=17, see r=16 but invert it. (Multiply R and the difference by -1 and if X=x, change it to X=x+1 and vice versa) For example:

**************************************************************************************************************************

If r=17, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is -4, the X=x+1, R=-4 and the fakes are in pile A.

If the difference is -8, the X=x+1, R=-4 and the fakes are in pile B.

If the difference is -12, the X=x+1, R=-4 and the fakes are in pile C.

**************************************************************************************************************************

If r=18, see r=15 but invert it.

r will not equal 19.

If r=20, see r=13 but invert it.

If r=21, see r=12 but invert it.

r will not equal 22.

If r=23, see r=10 but invert it.

If r=24, see r=9 but invert it.

If r=25, see r=8 but invert it.

r will not equal 26.

If r=27, see r=6 but invert it.

If r=28, see r=5 but invert it.

If r=29, see r=4 but invert it.

If r=30, see r=3 but invert it.

If r=31, see r=2 but invert it.

If r=32, see r=1 but invert it.

r will not equal 0.

The end. Phew! Anyone still with me? I didn't look over my work so there could be mistakes but the technique is sound.

Edit: Damn Bushido and his elegance!

Edited May 28, 2010 by Tuckleton

[Guest]

I could find out the way to estimate the actual weight of genuine coin and whether the fake coin is 5 gms less or 5 gms more than the genuine coin. That can be done in a single weighing. Take 1 coin each from each pile and weigh all 13 coins. If the total weight divided by 13 happens to contain decimal (0.615), then the fake coin is 5 gms less and if it contains (0.384), then the fake coin is 5 gms more. The weight of genuine coin is obtained by rounding off to higher integer in case of decimal (0.615) OR rounding off to lower integer (if the decimal is 0.384).

Having done this I am stuck now to identify the pile which contains fake coins. The limitation is that each pile contains max of 4 coins. Any body help ??

The counterfeit coins can be any real number between and including 5 and -5. If you weighed one coin from each pile and the result was 27, you would know that the genuine coins weigh 2, and the counterfeit weighs 3. But your going to be in big trouble figuring out the pile that contains the counterfeit coins on the next weighing.

[Guest]

Can Δ be 0?

Yes and no. If delta=0, the problem can be solved, but you would find that there are no counterfeit coins. This result would be counter to the description in the problem that there are counterfeit coins to find.

[bushindo]

Here goes,

Call the piles 1,2,3,4,5,6,7,8,9,0,A,B,C.

X is the normal coin weight and R is delta.

Put the following on the scale:

1 coin from each of 1,2 and 3

2 coins from each of 4,5 and 6

3 coins from each of 7,8,9 and 0

4 coins from each of A,B and C

Divide the total weight by 33 and leave the remainder. Call the quotient x and the remainder r.

If r=1, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 1, then X=x, R=1 and the fakes are in pile 1.

If the difference is 2, then X=x, R=1 and the fakes are in pile 2.

If the difference is 3, then X=x, R=1 and the fakes are in pile 3.

If r=2, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 4, 2 from 5 and 3 from 6 on the scale. Subtract 15x from the result.

If the difference is 1, the X=x, R=1 and the fakes are in pile 4.

If the difference is 2, the X=x, R=1 and the fakes are in pile 5.

If the difference is 3, the X=x, R=1 and the fakes are in pile 6.

If the difference is 4, the X=x, R=2 and the fakes are in pile 1.

If the difference is 6, the X=x, R=2 and the fakes are in pile 2.

If the difference is 8, the X=x, R=2 and the fakes are in pile 3.

If r=3, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 7, 2 from 8, 3 from 9, 4 from 10. Subtract 19x from the result.

If the difference is 1, the X=x, R=1 and the fakes are in pile 7.

If the difference is 2, the X=x, R=1 and the fakes are in pile 8.

If the difference is 3, the X=x, R=1 and the fakes are in pile 9.

If the difference is 4, the X=x, R=1 and the fakes are in pile 0.

If the difference is 6, the X=x, R=3 and the fakes are in pile 1.

If the difference is 9, the X=x, R=3 and the fakes are in pile 2.

If the difference is 12, the X=x, R=3 and the fakes are in pile 3.

If r=4, then put the following on the scale: 3 from 1, 4 from 2, 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 22x from the result.

If the difference is 1, the X=x, R=1 and the fakes are in pile A.

If the difference is 2, the X=x, R=1 and the fakes are in pile B.

If the difference is 3, the X=x, R=1 and the fakes are in pile C.

If the difference is 4, the X=x, R=2 and the fakes are in pile 4.

If the difference is 6, the X=x, R=2 and the fakes are in pile 5.

If the difference is 8, the X=x, R=2 and the fakes are in pile 6.

If the difference is 12, the X=x, R=3 and the fakes are in pile 1.

If the difference is 16, the X=x, R=3 and the fakes are in pile 2.

If the difference is 0, the X=x, R=3 and the fakes are in pile 3.

If r=5, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 5, the X=x, R=5 and the fakes are in pile 1.

If the difference is 10, the X=x, R=5 and the fakes are in pile 2.

If the difference is 15, the X=x, R=5 and the fakes are in pile 3.

If r=6, then put the following on the scale: 1 from 4, 3 from 5, 4 from 6, 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 18x from the result.

If the difference is 2, the X=x, R=2 and the fakes are in pile 7.

If the difference is 3, the X=x, R=3 and the fakes are in pile 4.

If the difference is 4, the X=x, R=2 and the fakes are in pile 8.

If the difference is 6, the X=x, R=2 and the fakes are in pile 9.

If the difference is 8, the X=x, R=2 and the fakes are in pile 0.

If the difference is 9, the X=x, R=3 and the fakes are in pile 5.

If the difference is 12, the X=x, R=3 and the fakes are in pile 6.

r will not equal 7.

If r=8, then put the following on the scale: 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 15x from the result.

If the difference is 2, the X=x, R=2 and the fakes are in pile A.

If the difference is 4, the X=x, R=2 and the fakes are in pile B.

If the difference is 6, the X=x, R=2 and the fakes are in pile C.

If the difference is 8, the X=x, R=4 and the fakes are in pile 4.

If the difference is 12, the X=x, R=4 and the fakes are in pile 5.

If the difference is 16, the X=x, R=4 and the fakes are in pile 6.

If r=9, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 3, the X=x, R=3 and the fakes are in pile 7.

If the difference is 6, the X=x, R=3 and the fakes are in pile 8.

If the difference is 9, the X=x, R=3 and the fakes are in pile 9.

If the difference is 12, the X=x, R=3 and the fakes are in pile 0.

If r=10, then put the following on the scale: 1 from 4, 2 from 5 and 3 from 6. Subtract 6x from the result.

If the difference is 5, the X=x, R=5 and the fakes are in pile 4.

If the difference is 10, the X=x, R=5 and the fakes are in pile 5.

If the difference is 15, the X=x, R=5 and the fakes are in pile 6.

r will not equal 11.

If r=12, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9, 4 from 0, 1 from A, 2 from B and 3 from C. Subtract 16x from the result.

If the difference is 3, the X=x, R=3 and the fakes are in pile A.

If the difference is 4, the X=x, R=4 and the fakes are in pile 7.

If the difference is 6, the X=x, R=3 and the fakes are in pile B.

If the difference is 8, the X=x, R=4 and the fakes are in pile 8.

If the difference is 9, the X=x, R=3 and the fakes are in pile C.

If the difference is 12, the X=x, R=4 and the fakes are in pile 9.

If the difference is 16, the X=x, R=4 and the fakes are in pile 0.

If the difference is , the X=x, R= and the fakes are in pile .

If r=13, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is -5, the X=x+1, R=-5 and the fakes are in pile A.

If the difference is -10, the X=x+1, R=-5 and the fakes are in pile B.

If the difference is -15, the X=x+1, R=-5 and the fakes are in pile C.

r will not equal 14.

If r=15, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 5, the X=x, R=5 and the fakes are in pile 7.

If the difference is 10, the X=x, R=5 and the fakes are in pile 8.

If the difference is 15, the X=x, R=5 and the fakes are in pile 9.

If the difference is 20, the X=x, R=5 and the fakes are in pile 0.

If r=16, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is 4, the X=x, R=4 and the fakes are in pile A.

If the difference is 8, the X=x, R=4 and the fakes are in pile B.

If the difference is 12, the X=x, R=4 and the fakes are in pile C.

If the difference is , the X=x, R= and the fakes are in pile .

If r=17, see r=16 but invert it. (Multiply R and the difference by -1 and if X=x, change it to X=x+1 and vice versa) For example:

**************************************************************************************************************************

If r=17, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is -4, the X=x+1, R=-4 and the fakes are in pile A.

If the difference is -8, the X=x+1, R=-4 and the fakes are in pile B.

If the difference is -12, the X=x+1, R=-4 and the fakes are in pile C.

**************************************************************************************************************************

If r=18, see r=15 but invert it.

r will not equal 19.

If r=20, see r=13 but invert it.

If r=21, see r=12 but invert it.

r will not equal 22.

If r=23, see r=10 but invert it.

If r=24, see r=9 but invert it.

If r=25, see r=8 but invert it.

r will not equal 26.

If r=27, see r=6 but invert it.

If r=28, see r=5 but invert it.

If r=29, see r=4 but invert it.

If r=30, see r=3 but invert it.

If r=31, see r=2 but invert it.

If r=32, see r=1 but invert it.

r will not equal 0.

The end. Phew! Anyone still with me? I didn't look over my work so there could be mistakes but the technique is sound.

Edit: Damn Bushido and his elegance!

Haha, that's nice complement indeed. You just made my day.

Edited May 28, 2010 by bushindo

[Guest]

Interesting problem. Here's my attempt

Label the piles from 1-13. We will only take coins from pile 1-12. For the first and second weighing, draw the following number of coin from each numbered pile and weigh those drawn coins. Note that the ratio of the first and second draw for every pile is a unique ratio. There are 12 such unique ratios.

     1st  2nd         ratio         

P 1  1     1      1.0000000

P 2  1     2      2.0000000

P 3  1     3      3.0000000

P 4  1     4      4.0000000

P 5  2     1      0.5000000

P 6  2     3      1.5000000

P 7  3     1      0.3333333

P 8  3     2      0.6666667

P 9  3     0      0.0000000

P 10 3     4      1.3333333

P 11 4     1      0.2500000

P 12 4     3      0.7500000

Now, for the first weighing, we will be putting 30 coins onto the register. If we get an integer as a total weight, then pile 13 is the odd pile and it is trivial to figure out delta. If the first weighting returns a real number (call it K₁), do the 2nd weighting as directed. There will be 25 coins for the second weighting and we'll get a second number (K₂). So, we have the following equation

30X - a * delta = K₁

25X - b * delta = K₂,

where a and b are the first and second draw number for some pile in the matrix above. X is either floor( K₁/30 ) or ceil( K₁/30 ). We can compute the ratio b/a by substituting these two possible values for X in

b/a = (K₂ - 25X )/ (K₁ - 30X ).

If we substitute the correct X, then the ratio b/a will be one of those in the matrix above, which will tell us which pile the odd coins are in. The rest of the parameters can then be computed.

not to quibble but there are only 28 coins for weight #1, but I think either amount would work as described.

[bushindo]

You're right. There are only 28 coins in the first weighing. Good catch. Thank you.

[plainglazed]

I hesitate to chime in, not only because I'm all too often obviously wrong here and probably especially so since I've only briefly scanned the responses, but also because I'm not sure how to read the OP. "There are 13 separate piles of coins. Each pile has 4 identical coins in it." Does that mean that in the pile with the odd coins, 4 are like all the rest and 9 are off by delta? And if so and you are picking only 1,2,3,or 4 coins, couldnt you pick all normal coins even from the pile containing odd ones? Or does this line, "The 1 pile of counterfeit coins look and feel identical to the genuine coins." mean all the coins in the counterfeit pile are off by delta? Like I said tho, am probably missing something obvious...

[Guest]

Interesting problem. Here's my attempt

Label the piles from 1-13. We will only take coins from pile 1-12. For the first and second weighing, draw the following number of coin from each numbered pile and weigh those drawn coins. Note that the ratio of the first and second draw for every pile is a unique ratio. There are 12 such unique ratios.

     1st  2nd         ratio         

P 1  1     1      1.0000000

P 2  1     2      2.0000000

P 3  1     3      3.0000000

P 4  1     4      4.0000000

P 5  2     1      0.5000000

P 6  2     3      1.5000000

P 7  3     1      0.3333333

P 8  3     2      0.6666667

P 9  3     0      0.0000000

P 10 3     4      1.3333333

P 11 4     1      0.2500000

P 12 4     3      0.7500000

Now, for the first weighing, we will be putting 30 coins onto the register. If we get an integer as a total weight, then pile 13 is the odd pile and it is trivial to figure out delta. If the first weighting returns a real number (call it K₁), do the 2nd weighting as directed. There will be 25 coins for the second weighting and we'll get a second number (K₂). So, we have the following equation

30X - a * delta = K₁

25X - b * delta = K₂,

where a and b are the first and second draw number for some pile in the matrix above. X is either floor( K₁/30 ) or ceil( K₁/30 ). We can compute the ratio b/a by substituting these two possible values for X in

b/a = (K₂ - 25X )/ (K₁ - 30X ).

If we substitute the correct X, then the ratio b/a will be one of those in the matrix above, which will tell us which pile the odd coins are in. The rest of the parameters can then be computed.

Bushido,

There's a minor error in your solution. You say that if the first weighing is an integer, then pile 13 has the counterfeits. You should say if the weighing is a multiple of 30, pile 13 contains the counterfeits. I busted a friend's solution that was very close to yours, but slightly different. I'll see if there could be a special case that will defeat your strategy. It may take me some time.

[Guest]

I hesitate to chime in, not only because I'm all too often obviously wrong here and probably especially so since I've only briefly scanned the responses, but also because I'm not sure how to read the OP. "There are 13 separate piles of coins. Each pile has 4 identical coins in it." Does that mean that in the pile with the odd coins, 4 are like all the rest and 9 are off by delta? And if so and you are picking only 1,2,3,or 4 coins, couldnt you pick all normal coins even from the pile containing odd ones? Or does this line, "The 1 pile of counterfeit coins look and feel identical to the genuine coins." mean all the coins in the counterfeit pile are off by delta? Like I said tho, am probably missing something obvious...

12 piles of genuine coins. In the one counterfeit pile, they all have the same delta.

[Guest]

You're right. There are only 28 coins in the first weighing. Good catch. Thank you.

Hmm. 28 might be easier to find a special case. I was having trouble with 30. My friend had 27 and 26 for his weighings. I'll work on it later.

[bushindo]

Hmm. 28 might be easier to find a special case. I was having trouble with 30. My friend had 27 and 26 for his weighings. I'll work on it later.

I was under the impression that delta is a random variable drawn from the range [ -5, 5], and hence it can not be a rational number as transcendental numbers are infinitely more dense. My algorithm was designed for work for that case. I see that delta is meant to be an arbitrarily chosen number from [ -5, 5].

Under this case, for certain rational values of delta, it is not possible to uniquely identify X, delta, and odd pile. Back to the drawing board.

[Guest]

Well in that case my method should work (once I fix a small oversight on my part, which I've just done.) Here's my updated solution, it may not be pretty but it works. Just pick a random pile, X and delta and try it out.

Here goes,

Call the piles 1,2,3,4,5,6,7,8,9,0,A,B,C.

X is the normal coin weight and d is delta.

Put the following on the scale:

1 coin from each of 1,2 and 3

2 coins from each of 4,5 and 6

3 coins from each of 7,8,9 and 0

4 coins from each of A,B and C

Divide the total weight by 33 and leave the remainder. Call the quotient x and the remainder r.

If r=1, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 1, then X=x, d=1 and the fakes are in pile 1.

If the difference is 2, then X=x, d=1 and the fakes are in pile 2.

If the difference is 3, then X=x, d=1 and the fakes are in pile 3.

If r=2, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 4, 2 from 5 and 3 from 6 on the scale. Subtract 15x from the result.

If the difference is 1, the X=x, d=1 and the fakes are in pile 4.

If the difference is 2, the X=x, d=1 and the fakes are in pile 5.

If the difference is 3, the X=x, d=1 and the fakes are in pile 6.

If the difference is 4, the X=x, d=2 and the fakes are in pile 1.

If the difference is 6, the X=x, d=2 and the fakes are in pile 2.

If the difference is 8, the X=x, d=2 and the fakes are in pile 3.

If r=3, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 7, 2 from 8, 3 from 9, 4 from 10. Subtract 19x from the result.

If the difference is 1, the X=x, d=1 and the fakes are in pile 7.

If the difference is 2, the X=x, d=1 and the fakes are in pile 8.

If the difference is 3, the X=x, d=1 and the fakes are in pile 9.

If the difference is 4, the X=x, d=1 and the fakes are in pile 0.

If the difference is 6, the X=x, d=3 and the fakes are in pile 1.

If the difference is 9, the X=x, d=3 and the fakes are in pile 2.

If the difference is 12, the X=x, d=3 and the fakes are in pile 3.

If r=4, then put the following on the scale: 3 from 1, 4 from 2, 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 22x from the result.

If the difference is 1, the X=x, d=1 and the fakes are in pile A.

If the difference is 2, the X=x, d=1 and the fakes are in pile B.

If the difference is 3, the X=x, d=1 and the fakes are in pile C.

If the difference is 4, the X=x, d=2 and the fakes are in pile 4.

If the difference is 6, the X=x, d=2 and the fakes are in pile 5.

If the difference is 8, the X=x, d=2 and the fakes are in pile 6.

If the difference is 12, the X=x, d=3 and the fakes are in pile 1.

If the difference is 16, the X=x, d=3 and the fakes are in pile 2.

If the difference is 0, the X=x, d=3 and the fakes are in pile 3.

If r=5, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile 1.

If the difference is 10, the X=x, d=5 and the fakes are in pile 2.

If the difference is 15, the X=x, d=5 and the fakes are in pile 3.

If r=6, then put the following on the scale: 1 from 4, 3 from 5, 4 from 6, 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 18x from the result.

If the difference is 2, the X=x, d=2 and the fakes are in pile 7.

If the difference is 3, the X=x, d=3 and the fakes are in pile 4.

If the difference is 4, the X=x, d=2 and the fakes are in pile 8.

If the difference is 6, the X=x, d=2 and the fakes are in pile 9.

If the difference is 8, the X=x, d=2 and the fakes are in pile 0.

If the difference is 9, the X=x, d=3 and the fakes are in pile 5.

If the difference is 12, the X=x, d=3 and the fakes are in pile 6.

r will not equal 7.

If r=8, then put the following on the scale: 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 15x from the result.

If the difference is 2, the X=x, d=2 and the fakes are in pile A.

If the difference is 4, the X=x, d=2 and the fakes are in pile B.

If the difference is 6, the X=x, d=2 and the fakes are in pile C.

If the difference is 8, the X=x, d=4 and the fakes are in pile 4.

If the difference is 12, the X=x, d=4 and the fakes are in pile 5.

If the difference is 16, the X=x, d=4 and the fakes are in pile 6.

If r=9, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 3, the X=x, d=3 and the fakes are in pile 7.

If the difference is 6, the X=x, d=3 and the fakes are in pile 8.

If the difference is 9, the X=x, d=3 and the fakes are in pile 9.

If the difference is 12, the X=x, d=3 and the fakes are in pile 0.

If r=10, then put the following on the scale: 1 from 4, 2 from 5 and 3 from 6. Subtract 6x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile 4.

If the difference is 10, the X=x, d=5 and the fakes are in pile 5.

If the difference is 15, the X=x, d=5 and the fakes are in pile 6.

r will not equal 11.

If r=12, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9, 4 from 0, 1 from A, 2 from B and 3 from C. Subtract 16x from the result.

If the difference is 3, the X=x, d=3 and the fakes are in pile A.

If the difference is 4, the X=x, d=4 and the fakes are in pile 7.

If the difference is 6, the X=x, d=3 and the fakes are in pile B.

If the difference is 8, the X=x, d=4 and the fakes are in pile 8.

If the difference is 9, the X=x, d=3 and the fakes are in pile C.

If the difference is 12, the X=x, d=4 and the fakes are in pile 9.

If the difference is 16, the X=x, d=4 and the fakes are in pile 0.

If r=13, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile A.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile B.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile C.

r will not equal 14.

If r=15, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile 7.

If the difference is 10, the X=x, d=5 and the fakes are in pile 8.

If the difference is 15, the X=x, d=5 and the fakes are in pile 9.

If the difference is 20, the X=x, d=5 and the fakes are in pile 0.

If r=16, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is 4, the X=x, d=4 and the fakes are in pile A.

If the difference is 8, the X=x, d=4 and the fakes are in pile B.

If the difference is 12, the X=x, d=4 and the fakes are in pile C.

If r=17, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6(x+1) from the result.

If the difference is -4, the X=x+1, d=-4 and the fakes are in pile A.

If the difference is -8, the X=x+1, d=-4 and the fakes are in pile B.

If the difference is -12, the X=x+1, d=-4 and the fakes are in pile C.

If r=18, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile 7.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile 8.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile 9.

If the difference is -20, the X=x+1, d=-5 and the fakes are in pile 0.

r will not equal 19.

If r=20, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile A.

If the difference is 10, the X=x, d=5 and the fakes are in pile B.

If the difference is 15, the X=x, d=5 and the fakes are in pile C.

If r=21, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9, 4 from 0, 1 from A, 2 from B and 3 from C. Subtract 16(x+1) from the result.

If the difference is -3, the X=x+1, d=-3 and the fakes are in pile A.

If the difference is -4, the X=x+1, d=-4 and the fakes are in pile 7.

If the difference is -6, the X=x+1, d=-3 and the fakes are in pile B.

If the difference is -8, the X=x+1, d=-4 and the fakes are in pile 8.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile C.

If the difference is -12, the X=x+1, d=-4 and the fakes are in pile 9.

If the difference is -16, the X=x+1, d=-4 and the fakes are in pile 0.

r will not equal 22.

If r=23, then put the following on the scale: 1 from 4, 2 from 5 and 3 from 6. Subtract 6(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile 4.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile 5.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile 6.

If r=24, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10(x+1) from the result.

If the difference is -3, the X=x+1, d=-3 and the fakes are in pile 7.

If the difference is -6, the X=x+1, d=-3 and the fakes are in pile 8.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile 9.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 0.

If r=25, then put the following on the scale: 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 15(x+1) from the result.

If the difference is -2, the X=x+1, d=-2 and the fakes are in pile A.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile B.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile C.

If the difference is -8, the X=x+1, d=-4 and the fakes are in pile 4.

If the difference is -12, the X=x+1, d=-4 and the fakes are in pile 5.

If the difference is -16, the X=x+1, d=-4 and the fakes are in pile 6.

r will not equal 26.

If r=27, then put the following on the scale: 1 from 4, 3 from 5, 4 from 6, 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 18(x+1) from the result.

If the difference is -2, the X=x+1, d=-2 and the fakes are in pile 7.

If the difference is -3, the X=x+1, d=-3 and the fakes are in pile 4.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile 8.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile 9.

If the difference is -8, the X=x+1, d=-2 and the fakes are in pile 0.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile 5.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 6.

If r=28, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile 1.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile 2.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile 3.

If r=29, see then put the following on the scale: 3 from 1, 4 from 2, 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 22(x+1) from the result.

If the difference is -1, the X=x+1, d=-1 and the fakes are in pile A.

If the difference is -2, the X=x+1, d=-1 and the fakes are in pile B.

If the difference is -3, the X=x+1, d=-1 and the fakes are in pile C.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile 4.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile 5.

If the difference is -8, the X=x+1, d=-2 and the fakes are in pile 6.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 1.

If the difference is -16, the X=x+1, d=-3 and the fakes are in pile 2.

If the difference is 0, the X=x+1, d=-3 and the fakes are in pile 3.

If r=30, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 7, 2 from 8, 3 from 9, 4 from 10. Subtract 19(x+1) from the result.

If the difference is -1, the X=x+1, d=-1 and the fakes are in pile 7.

If the difference is -2, the X=x+1, d=-1 and the fakes are in pile 8.

If the difference is -3, the X=x+1, d=-1 and the fakes are in pile 9.

If the difference is -4, the X=x+1, d=-1 and the fakes are in pile 0.

If the difference is -6, the X=x+1, d=-3 and the fakes are in pile 1.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile 2.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 3.

If r=31, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 4, 2 from 5 and 3 from 6 on the scale. Subtract 15(x+1) from the result.

If the difference is -1, the X=x+1, d=-1 and the fakes are in pile 4.

If the difference is -2, the X=x+1, d=-1 and the fakes are in pile 5.

If the difference is -3, the X=x+1, d=-1 and the fakes are in pile 6.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile 1.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile 2.

If the difference is -8, the X=x+1, d=-2 and the fakes are in pile 3.

If r=32, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6(x+1) from the result.

If the difference is -1, then X=x+1, d=-1 and the fakes are in pile 1.

If the difference is -2, then X=x+1, d=-1 and the fakes are in pile 2.

If the difference is -3, then X=x+1, d=-1 and the fakes are in pile 3.

r will not equal 0.

The end. Phew! Anyone still with me? I didn't look over my work so there could be mistakes but the technique is sound.

[bushindo]

Well in that case my method should work (once I fix a small oversight on my part, which I've just done.) Here's my updated solution, it may not be pretty but it works. Just pick a random pile, X and delta and try it out.

Here goes,

Call the piles 1,2,3,4,5,6,7,8,9,0,A,B,C.

X is the normal coin weight and d is delta.

Put the following on the scale:

1 coin from each of 1,2 and 3

2 coins from each of 4,5 and 6

3 coins from each of 7,8,9 and 0

4 coins from each of A,B and C

Divide the total weight by 33 and leave the remainder. Call the quotient x and the remainder r.

If r=1, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 1, then X=x, d=1 and the fakes are in pile 1.

If the difference is 2, then X=x, d=1 and the fakes are in pile 2.

If the difference is 3, then X=x, d=1 and the fakes are in pile 3.

If r=2, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 4, 2 from 5 and 3 from 6 on the scale. Subtract 15x from the result.

If the difference is 1, the X=x, d=1 and the fakes are in pile 4.

If the difference is 2, the X=x, d=1 and the fakes are in pile 5.

If the difference is 3, the X=x, d=1 and the fakes are in pile 6.

If the difference is 4, the X=x, d=2 and the fakes are in pile 1.

If the difference is 6, the X=x, d=2 and the fakes are in pile 2.

If the difference is 8, the X=x, d=2 and the fakes are in pile 3.

If r=3, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 7, 2 from 8, 3 from 9, 4 from 10. Subtract 19x from the result.

If the difference is 1, the X=x, d=1 and the fakes are in pile 7.

If the difference is 2, the X=x, d=1 and the fakes are in pile 8.

If the difference is 3, the X=x, d=1 and the fakes are in pile 9.

If the difference is 4, the X=x, d=1 and the fakes are in pile 0.

If the difference is 6, the X=x, d=3 and the fakes are in pile 1.

If the difference is 9, the X=x, d=3 and the fakes are in pile 2.

If the difference is 12, the X=x, d=3 and the fakes are in pile 3.

If r=4, then put the following on the scale: 3 from 1, 4 from 2, 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 22x from the result.

If the difference is 1, the X=x, d=1 and the fakes are in pile A.

If the difference is 2, the X=x, d=1 and the fakes are in pile B.

If the difference is 3, the X=x, d=1 and the fakes are in pile C.

If the difference is 4, the X=x, d=2 and the fakes are in pile 4.

If the difference is 6, the X=x, d=2 and the fakes are in pile 5.

If the difference is 8, the X=x, d=2 and the fakes are in pile 6.

If the difference is 12, the X=x, d=3 and the fakes are in pile 1.

If the difference is 16, the X=x, d=3 and the fakes are in pile 2.

If the difference is 0, the X=x, d=3 and the fakes are in pile 3.

If r=5, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile 1.

If the difference is 10, the X=x, d=5 and the fakes are in pile 2.

If the difference is 15, the X=x, d=5 and the fakes are in pile 3.

If r=6, then put the following on the scale: 1 from 4, 3 from 5, 4 from 6, 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 18x from the result.

If the difference is 2, the X=x, d=2 and the fakes are in pile 7.

If the difference is 3, the X=x, d=3 and the fakes are in pile 4.

If the difference is 4, the X=x, d=2 and the fakes are in pile 8.

If the difference is 6, the X=x, d=2 and the fakes are in pile 9.

If the difference is 8, the X=x, d=2 and the fakes are in pile 0.

If the difference is 9, the X=x, d=3 and the fakes are in pile 5.

If the difference is 12, the X=x, d=3 and the fakes are in pile 6.

r will not equal 7.

If r=8, then put the following on the scale: 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 15x from the result.

If the difference is 2, the X=x, d=2 and the fakes are in pile A.

If the difference is 4, the X=x, d=2 and the fakes are in pile B.

If the difference is 6, the X=x, d=2 and the fakes are in pile C.

If the difference is 8, the X=x, d=4 and the fakes are in pile 4.

If the difference is 12, the X=x, d=4 and the fakes are in pile 5.

If the difference is 16, the X=x, d=4 and the fakes are in pile 6.

If r=9, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 3, the X=x, d=3 and the fakes are in pile 7.

If the difference is 6, the X=x, d=3 and the fakes are in pile 8.

If the difference is 9, the X=x, d=3 and the fakes are in pile 9.

If the difference is 12, the X=x, d=3 and the fakes are in pile 0.

If r=10, then put the following on the scale: 1 from 4, 2 from 5 and 3 from 6. Subtract 6x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile 4.

If the difference is 10, the X=x, d=5 and the fakes are in pile 5.

If the difference is 15, the X=x, d=5 and the fakes are in pile 6.

r will not equal 11.

If r=12, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9, 4 from 0, 1 from A, 2 from B and 3 from C. Subtract 16x from the result.

If the difference is 3, the X=x, d=3 and the fakes are in pile A.

If the difference is 4, the X=x, d=4 and the fakes are in pile 7.

If the difference is 6, the X=x, d=3 and the fakes are in pile B.

If the difference is 8, the X=x, d=4 and the fakes are in pile 8.

If the difference is 9, the X=x, d=3 and the fakes are in pile C.

If the difference is 12, the X=x, d=4 and the fakes are in pile 9.

If the difference is 16, the X=x, d=4 and the fakes are in pile 0.

If r=13, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile A.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile B.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile C.

r will not equal 14.

If r=15, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile 7.

If the difference is 10, the X=x, d=5 and the fakes are in pile 8.

If the difference is 15, the X=x, d=5 and the fakes are in pile 9.

If the difference is 20, the X=x, d=5 and the fakes are in pile 0.

If r=16, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is 4, the X=x, d=4 and the fakes are in pile A.

If the difference is 8, the X=x, d=4 and the fakes are in pile B.

If the difference is 12, the X=x, d=4 and the fakes are in pile C.

If r=17, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6(x+1) from the result.

If the difference is -4, the X=x+1, d=-4 and the fakes are in pile A.

If the difference is -8, the X=x+1, d=-4 and the fakes are in pile B.

If the difference is -12, the X=x+1, d=-4 and the fakes are in pile C.

If r=18, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile 7.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile 8.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile 9.

If the difference is -20, the X=x+1, d=-5 and the fakes are in pile 0.

r will not equal 19.

If r=20, then put the following on the scale: 1 from A, 2 from B and 3 from C. Subtract 6x from the result.

If the difference is 5, the X=x, d=5 and the fakes are in pile A.

If the difference is 10, the X=x, d=5 and the fakes are in pile B.

If the difference is 15, the X=x, d=5 and the fakes are in pile C.

If r=21, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9, 4 from 0, 1 from A, 2 from B and 3 from C. Subtract 16(x+1) from the result.

If the difference is -3, the X=x+1, d=-3 and the fakes are in pile A.

If the difference is -4, the X=x+1, d=-4 and the fakes are in pile 7.

If the difference is -6, the X=x+1, d=-3 and the fakes are in pile B.

If the difference is -8, the X=x+1, d=-4 and the fakes are in pile 8.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile C.

If the difference is -12, the X=x+1, d=-4 and the fakes are in pile 9.

If the difference is -16, the X=x+1, d=-4 and the fakes are in pile 0.

r will not equal 22.

If r=23, then put the following on the scale: 1 from 4, 2 from 5 and 3 from 6. Subtract 6(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile 4.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile 5.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile 6.

If r=24, then put the following on the scale: 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 10(x+1) from the result.

If the difference is -3, the X=x+1, d=-3 and the fakes are in pile 7.

If the difference is -6, the X=x+1, d=-3 and the fakes are in pile 8.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile 9.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 0.

If r=25, then put the following on the scale: 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 15(x+1) from the result.

If the difference is -2, the X=x+1, d=-2 and the fakes are in pile A.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile B.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile C.

If the difference is -8, the X=x+1, d=-4 and the fakes are in pile 4.

If the difference is -12, the X=x+1, d=-4 and the fakes are in pile 5.

If the difference is -16, the X=x+1, d=-4 and the fakes are in pile 6.

r will not equal 26.

If r=27, then put the following on the scale: 1 from 4, 3 from 5, 4 from 6, 1 from 7, 2 from 8, 3 from 9 and 4 from 0. Subtract 18(x+1) from the result.

If the difference is -2, the X=x+1, d=-2 and the fakes are in pile 7.

If the difference is -3, the X=x+1, d=-3 and the fakes are in pile 4.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile 8.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile 9.

If the difference is -8, the X=x+1, d=-2 and the fakes are in pile 0.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile 5.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 6.

If r=28, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6(x+1) from the result.

If the difference is -5, the X=x+1, d=-5 and the fakes are in pile 1.

If the difference is -10, the X=x+1, d=-5 and the fakes are in pile 2.

If the difference is -15, the X=x+1, d=-5 and the fakes are in pile 3.

If r=29, see then put the following on the scale: 3 from 1, 4 from 2, 2 from 4, 3 from 5, 4 from 6, 1 from A, 2 from B and 3 from C. Subtract 22(x+1) from the result.

If the difference is -1, the X=x+1, d=-1 and the fakes are in pile A.

If the difference is -2, the X=x+1, d=-1 and the fakes are in pile B.

If the difference is -3, the X=x+1, d=-1 and the fakes are in pile C.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile 4.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile 5.

If the difference is -8, the X=x+1, d=-2 and the fakes are in pile 6.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 1.

If the difference is -16, the X=x+1, d=-3 and the fakes are in pile 2.

If the difference is 0, the X=x+1, d=-3 and the fakes are in pile 3.

If r=30, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 7, 2 from 8, 3 from 9, 4 from 10. Subtract 19(x+1) from the result.

If the difference is -1, the X=x+1, d=-1 and the fakes are in pile 7.

If the difference is -2, the X=x+1, d=-1 and the fakes are in pile 8.

If the difference is -3, the X=x+1, d=-1 and the fakes are in pile 9.

If the difference is -4, the X=x+1, d=-1 and the fakes are in pile 0.

If the difference is -6, the X=x+1, d=-3 and the fakes are in pile 1.

If the difference is -9, the X=x+1, d=-3 and the fakes are in pile 2.

If the difference is -12, the X=x+1, d=-3 and the fakes are in pile 3.

If r=31, then put the following on the scale: 2 from 1, 3 from 2, 4 from 3, 1 from 4, 2 from 5 and 3 from 6 on the scale. Subtract 15(x+1) from the result.

If the difference is -1, the X=x+1, d=-1 and the fakes are in pile 4.

If the difference is -2, the X=x+1, d=-1 and the fakes are in pile 5.

If the difference is -3, the X=x+1, d=-1 and the fakes are in pile 6.

If the difference is -4, the X=x+1, d=-2 and the fakes are in pile 1.

If the difference is -6, the X=x+1, d=-2 and the fakes are in pile 2.

If the difference is -8, the X=x+1, d=-2 and the fakes are in pile 3.

If r=32, then put the following on the scale: 1 from 1, 2 from 2 and 3 from 3. Subtract 6(x+1) from the result.

If the difference is -1, then X=x+1, d=-1 and the fakes are in pile 1.

If the difference is -2, then X=x+1, d=-1 and the fakes are in pile 2.

If the difference is -3, then X=x+1, d=-1 and the fakes are in pile 3.

r will not equal 0.

The end. Phew! Anyone still with me? I didn't look over my work so there could be mistakes but the technique is sound.

Some comments

This method only seems to return integer values for delta, but delta is any real number on [-5, 5]. There are identifiability issues for certain values of delta. For instance, the four situations (delta = 1, pile = 1), (delta = 1/2, pile 4), (delta = 1/3, pile = 7), and (delta = 1/4, pile = A ) all give the same remainder 1.

[bushindo]

Hmm. 28 might be easier to find a special case. I was having trouble with 30. My friend had 27 and 26 for his weighings. I'll work on it later.

Okay, second try. This approach basically refines the initial estimates of X so that we don't have to worry about identifiability issue.

Label the piles from 1-13. We will only take coins from pile 1-12. For the first and second weighing, draw the following number of coin from each numbered pile and weigh those drawn coins. Note that the ratio of the first and second draw for every pile is a unique ratio. There are 12 such unique ratios.

   first second ratio

1      4      1  0.25

2      4      2  0.50

3      4      3  0.75

4      4      0  0.00

5      3      1  0.33

6      3      2  0.67

7      3      3  1.00

8      3      4  1.33

9      2      3  1.50

10     1      2  2.00

11     1      3  3.00

12     1      4  4.00

Now, for the first weighing, we will be putting 33 coins onto the register. If we get an multiple of 33 as a total weight, then pile 13 is the odd pile and it is trivial to figure out delta. Otherwise, record the scale reading (call it K₁), do the 2nd weighing as directed. There will be 28 coins for the second weighing and we'll get a second number (K₂).

So, we have the following equation

33X - a * delta = K₁

28X - b * delta = K₂,

where a and b are the first and second draw number for some pile in the matrix above. We now estimate X in order to substitute into the equation above. There are two possible values for X, which are floor( (K₁/33 ) and ceil(K₁/33 ).

Let r₁ = (K₁/33 ) mod 1. That is, let r₁ be the part after the decimal point of the average found by dividing the scale reading of the first weighing by 33. Let r₂ = (K₂/28 ) mod 1. In the trivial case that r₁ = 0 and r₂ = 0, then delta = 0. Otherwise,

if r₁ > 20/33 OR r₂ > 20/28, let X = ceil( (K₁/33 ).

if r₁ < 13/33 OR r₂ < 8/28, let X = floor( (K₁/33 ).

If none of the two condition above applies,

if r₂/r₁ * 28/33 is equal to one of the ratios in the matrix above, let X = floor( (K₁/33 ).

Otherwise, let X = ceil( (K₁/33 ).

We can compute the ratio b/a by substituting our X from above into

b/a = (K₂ - 28X )/ (K₁ - 33X ).

When we substitute X, then the ratio b/a will be one of those in the matrix above, which will tell us which pile the odd coins are in. In the rare case that b/a isn't one of the 12 unique ratios, substitute the other value for X. The rest of the parameters can then be computed.

Edited May 29, 2010 by bushindo

[bushindo]

Okay, third try. Please ignore the post above. I refined the approach a bit, and now I believe this one is correct. It will always give the correct X, delta, and pile number.

For the first and second weighing, draw the following number of coin from each numbered pile and weigh those drawn coins.

   first second ratio

1      4      1 0.250

2      4      3 0.750

3      1      0 0.000

4      3      1 0.333

5      3      2 0.667

6      3      4 1.333

7      2      1 0.500

8      2      2 1.000

9      2      3 1.500

10     2      4 2.000

11     1      4 4.000

12     1      5 5.000

Now, for the first weighing, we will be putting 28 coins onto the register. If we get an multiple of 28 as a total weight, then pile 13 is the odd pile and it is trivial to figure out delta. Otherwise, record the scale reading (call it K₁), do the 2nd weighing as directed. There will be 30 coins for the second weighing and we'll get a second number (K₂).

So, we have the following equation

33X - a * delta = K₁

28X - b * delta = K₂,

where a and b are the first and second draw number for some pile in the matrix above. We now estimate X in order to substitute into the equation above. There are two possible values for X, which are floor( (K₁/28 ) and ceil(K₁/28 ). We need to choose the correct one, which we will do in the following.

Let r₁ = (K₁/28 ) mod 1. That is, let r₁ be the part after the decimal point of the average found by dividing the scale reading of the first weighing by 28. Let r₂ = (K₂/30 ) mod 1. In the trivial case that r₁ = 0 and r₂ = 0, then delta = 0. Otherwise,

if r₁ > 20/28 OR r₂ > 20/30, let X = ceil( (K₁/28 ).

if r₁ < 8/28 OR r₂ < 10/30, let X = floor( (K₁/28 ).

If none of the two conditions above applies, then apply the following

if r₂/r₁ * 30/28 is equal to one of the ratios in the matrix above, let X = floor( (K₁/28 ).

Otherwise, let X = ceil( (K₁/28 ).

We can compute the ratio b/a by substituting our X from above into

b/a = (K₂ - 30X )/ (K₁ - 28X ).

When we substitute X, then the ratio b/a will be one of those in the matrix above, which will tell us which pile the odd coins are in. The rest of the parameters can then be computed.

Edited May 29, 2010 by bushindo

[bushindo]

Okay, third try. Please ignore the post above. I refined the approach a bit, and now I believe this one is correct. It will always give the correct X, delta, and pile number.

Please ignore the above posts. Unfortunately, there were grave errors in these approaches. It works generally, but there are 2-4 special pairs of (delta,pile number) that can not be uniquely identified. If there is a solution for this problem, I'd like to see it.

[Guest]

Please ignore the above posts. Unfortunately, there were grave errors in these approaches. It works generally, but there are 2-4 special pairs of (delta,pile number) that can not be uniquely identified. If there is a solution for this problem, I'd like to see it.

I'm ging to include two spoilers in this reply. The first will make a few observations about the problem and should provide some hints to the solution. The second will be my solution. Though there are others.

First, Bushido basically had it with his first attempt. He just can't quite be so "elegant." You're allowed to decide you second weighing after seeing the result of the first weighing. So if he has a slightly different weighing in reserve, he use that one if one of the few special cases are possible based on the first weighing. I don't believe there is a way to solve the problem for all cases without taking the result of the first weighing into account.

Second, the key number in this problem is 41. Since the delta can't be greater than 5 or less than -5 and there are only 4 counterfeit coins, if you can weigh 41 coins, then the result has to yield X and the sign of delta, at a minimum.

I will define one other concept that will help you understand the solution. DeltaSum is defined as the difference between the weighing, if all the coins were genuine and the actual weight. For example, if we were to weigh 41 coins and the result was 430, we would know that X=10 and delta must be 5 and the counterfeits are in a pile where we weighed 4 coins. We know this because the DeltaSum is 20 (the difference between 410 and the actual weight).

Another example, if we weigh 35 coins and the result is 370, the DeltaSum is either 20 or -15 and X is either 10 or 11. We also know that the counterfeits must be in a pile where we weighed either 3 or 4 coins. You can see by these examples, the power of weighing 41 coins.

One last concept. I will assume for my solution that X is either 10 or 11. Since any weighing will narrow the possible X's to two possible integers, this is a legal cheat.

This solution will be long and I'm going to use specific examples to illustrate various types of results. But these methods will work for all possible answers.

For my first weighing I will take 4 coins from piles 1, 2, 3,and 4. I will take 3 coins from piles 5, 6, and 7. I will take 1 coin from piles 8, 9 and 10. I will take 2 coins from piles 11 and 12. And I will take zero coins from pile 13. We are using 33 coins for this weighing. For this solution, I'm assuming X either equals 10 or 11, but my method will cover any other possible X. The weighing will yield three very different cases:

1. The easiest case is when the result is 330 (or some multiple of 33). We know the counterfeits are in the 13th pile, and X=10. We can simply weigh one coin from pile 13 and get Delta.

2. If the first weighing yields a result where the DeltaSum is less than 13, we will know what X is and the sign of Delta, and we eliminate pile 13. For example, if the weight is 340, X must be 10 and the delta must be positive. If it's 355, X must be 11 and the delta is negative. We would also be able to eliminate all the piles where we weighed 1 coin. But let's look at the hardest case for this type of result.

If the yield is 334, we will know that X=10 and the sign is positive. But all the piles are possible for the counterfeits (except for 13). We also know that Delta is 1, if the counterfeits are in pile 1, 2, 3, or 4. We know that Delta is 4/3 if the counterfeits are in piles 5, 6, or 7. We know the Delta is 2 the counterfeits are in piles 8, 9, or 10. And we know the Delta is 4 if the counterfeits are in piles 11 or 12. Luckily, we no longer need to weigh 41 coins, as we've answered most of the questions.

My second weighing will look like this:

Pile 1- 4 coins

Pile 2- 3 coins

Pile 3- 2 coins

Pile 4- 1 coins

Pile 5- 4 coins

Pile 6- 2 coins

Pile 7- 1 coins

Pile 8- 4 coins

Pile 9- 3 coins

Pile 10- 0 coins

Pile 11- 4 coins

Pile 12- 3 coins

We are using 31 coins. Here are the possible results:

314- Pile 1 has the counterfeits with Delta 1

313- Pile 2 with Delta 1

312- Pile 3 with Delta 1

311- Pile 4 with Delta 1

315+1/3- Pile 5 with Delta 4/3

313+2/3- Pile 6 with Delta 4/3

311+1/3- Pile 7 with Delta 4/3

318- Pile 8 with Delta 2

316- Pile 9 with Delta 2

310- Pile 10 with Delta 2 (you'll see why I had to use this pile for a zero second weighing, when you see part 3)

326- Pile 11 with Delta 4

322- Pile 12 with Delta 4

I think this method will work for all case where the DeltaSum is relatively small.

3. If the first weighing is between 343 and 350, we won't know X or the sign of Delta. But we can eliminate any pile where we weighed less than 3 coins (piles 8-13). This little detail caused me great headaches, when I weighed 4 coins from 5 piles. I figured I could weigh zero coins from one of those piles on my second weighing, but then I would come up short on my goal of getting to 41 coins.

Let's say the first weighing yields 350. X is either 10 or 11 and the Delta is either 5 (in which case we know the counterfeits are in piles 1, 2, 3 or 4) or the Delta is -13/4 (in which case the counterfeits are in piles 1, 2, 3 or 4). Or the Delta is -13/3 (and the counterfeits are in piles 5, 6 or 7). This is what my second weighing will look like:

Pile 1- 4

Pile 2- 3

Pile 3- 2

Pile 4- 1

Pile 5- 4

Pile 6- 2

Pile 7- 1

Pile 8- 4

Pile 9- 4

Pile 10- 4

Pile 11- 4

Pile 12- 4

Pile 13- 4

Exactly 41 coins!

Here are the possible results for weighing 2:

430- Pile 1 with Delta 5 and X=10

425- Pile 2 with Delta 5 and X=10

420- Pile 3 with Delta 5 and X=10

415- Pile 4 with Delta 5 and X=10

438- Pile 1 with Delta -13/4 and X=11

451.25- Pile 2 with Delta -13/4 and X=11

454.5- Pile 3 with Delta -13/4 and X=11

457.75- Pile 4 with Delta -13/4 and X=11

433+2/3- Pile 5 with Delta -13/3 and X=11

442+1/3- Pile 6 with Delta -13/3 and X=11

446+2/3- Pile 7 with Delta -13/3 and X=11

There are an infinite number of possible variables for X and Delta, but I think my methods, when used for the right cases, will cover all of them.

Edited May 30, 2010 by bonanova
1 coin from piles 8, 9

[Guest]

I'm ging to include two spoilers in this reply. The first will make a few observations about the problem and should provide some hints to the solution. The second will be my solution. Though there are others.

First, Bushido basically had it with his first attempt. He just can't quite be so "elegant." You're allowed to decide you second weighing after seeing the result of the first weighing. So if he has a slightly different weighing in reserve, he use that one if one of the few special cases are possible based on the first weighing. I don't believe there is a way to solve the problem for all cases without taking the result of the first weighing into account.

Second, the key number in this problem is 41. Since the delta can't be greater than 5 or less than -5 and there are only 4 counterfeit coins, if you can weigh 41 coins, then the result has to yield X and the sign of delta, at a minimum.

I will define one other concept that will help you understand the solution. DeltaSum is defined as the difference between the weighing, if all the coins were genuine and the actual weight. For example, if we were to weigh 41 coins and the result was 430, we would know that X=10 and delta must be 5 and the counterfeits are in a pile where we weighed 4 coins. We know this because the DeltaSum is 20 (the difference between 410 and the actual weight).

Another example, if we weigh 35 coins and the result is 370, the DeltaSum is either 20 or -15 and X is either 10 or 11. We also know that the counterfeits must be in a pile where we weighed either 3 or 4 coins. You can see by these examples, the power of weighing 41 coins.

One last concept. I will assume for my solution that X is either 10 or 11. Since any weighing will narrow the possible X's to two possible integers, this is a legal cheat.

This solution will be long and I'm going to use specific examples to illustrate various types of results. But these methods will work for all possible answers.

For my first weighing I will take 4 coins from piles 1, 2, 3,and 4. I will take 3 coins from piles 5, 6, and 7. I will take 2 coins from piles 8, 9 and 10. I will take 1 coin from piles 11 and 12. And I will take zero coins from pile 13. We are using 33 coins for this weighing. For this solution, I'm assuming X either equals 10 or 11, but my method will cover any other possible X. The weighing will yield three very different cases:

1. The easiest case is when the result is 330 (or some multiple of 33). We know the counterfeits are in the 13th pile, and X=10. We can simply weigh one coin from pile 13 and get Delta.

2. If the first weighing yields a result where the DeltaSum is less than 13, we will know what X is and the sign of Delta, and we eliminate pile 13. For example, if the weight is 340, X must be 10 and the delta must be positive. If it's 355, X must be 11 and the delta is negative. We would also be able to eliminate all the piles where we weighed 1 coin. But let's look at the hardest case for this type of result.

If the yield is 334, we will know that X=10 and the sign is positive. But all the piles are possible for the counterfeits (except for 13). We also know that Delta is 1, if the counterfeits are in pile 1, 2, 3, or 4. We know that Delta is 4/3 if the counterfeits are in piles 5, 6, or 7. We know the Delta is 2 the counterfeits are in piles 8, 9, or 10. And we know the Delta is 4 if the counterfeits are in piles 11 or 12. Luckily, we no longer need to weigh 41 coins, as we've answered most of the questions.

My second weighing will look like this:

Pile 1- 4 coins

Pile 2- 3 coins

Pile 3- 2 coins

Pile 4- 1 coins

Pile 5- 4 coins

Pile 6- 2 coins

Pile 7- 1 coins

Pile 8- 4 coins

Pile 9- 3 coins

Pile 10- 0 coins

Pile 11- 4 coins

Pile 12- 3 coins

We are using 31 coins. Here are the possible results:

314- Pile 1 has the counterfeits with Delta 1

313- Pile 2 with Delta 1

312- Pile 3 with Delta 1

311- Pile 4 with Delta 1

315+1/3- Pile 5 with Delta 4/3

313+2/3- Pile 6 with Delta 4/3

311+1/3- Pile 7 with Delta 4/3

318- Pile 8 with Delta 2

316- Pile 9 with Delta 2

310- Pile 10 with Delta 2 (you'll see why I had to use this pile for a zero second weighing, when you see part 3)

326- Pile 11 with Delta 4

322- Pile 12 with Delta 4

I think this method will work for all case where the DeltaSum is relatively small.

3. If the first weighing is between 343 and 350, we won't know X or the sign of Delta. But we can eliminate any pile where we weighed less than 3 coins (piles 8-13). This little detail caused me great headaches, when I weighed 4 coins from 5 piles. I figured I could weigh zero coins from one of those piles on my second weighing, but then I would come up short on my goal of getting to 41 coins.

Let's say the first weighing yields 350. X is either 10 or 11 and the Delta is either 5 (in which case we know the counterfeits are in piles 1, 2, 3 or 4) or the Delta is -13/4 (in which case the counterfeits are in piles 1, 2, 3 or 4). Or the Delta is -13/3 (and the counterfeits are in piles 5, 6 or 7). This is what my second weighing will look like:

Pile 1- 4

Pile 2- 3

Pile 3- 2

Pile 4- 1

Pile 5- 4

Pile 6- 2

Pile 7- 1

Pile 8- 4

Pile 9- 4

Pile 10- 4

Pile 11- 4

Pile 12- 4

Pile 13- 4

Exactly 41 coins!

Here are the possible results for weighing 2:

430- Pile 1 with Delta 5 and X=10

425- Pile 2 with Delta 5 and X=10

420- Pile 3 with Delta 5 and X=10

415- Pile 4 with Delta 5 and X=10

438- Pile 1 with Delta -13/4 and X=11

451.25- Pile 2 with Delta -13/4 and X=11

454.5- Pile 3 with Delta -13/4 and X=11

457.75- Pile 4 with Delta -13/4 and X=11

433+2/3- Pile 5 with Delta -13/3 and X=11

442+1/3- Pile 6 with Delta -13/3 and X=11

446+2/3- Pile 7 with Delta -13/3 and X=11

There are an infinite number of possible variables for X and Delta, but I think my methods, when used for the right cases, will cover all of them.

How do I edit a post? If someone else can, but I can't, I'd appreciate a courtesy edit to my solution. I mistakenly said I'd take 2 coins from piles 11 and 12. I meant to say 1 coin from 11 and 12.

Thanks

Edited May 30, 2010 by bonanova
Change 2 to 1 for piles 11 and 12 - OP request

[bonanova]

Lovely.

You have a trivial typo "I will take 2 1 coins from piles 11 and 12."

And I'll happily make the change at your request.

- bn

[bonanova]

Yes and no. If delta=0, the problem can be solved, but you would find that there are no counterfeit coins.

This result would be counter to the description in the problem that there are counterfeit coins to find.

Wouldn't it be the case that if Delta=0, the counterfeit coins cannot be identified?

OP states there are counterfeits; their weight is within limits that include the weight of genuine coins.

Although it could be determined that Delta = 0; weighing would seem powerless to identify the counterfeit pile.

That is, the counterfeits could be struck from material of density equal to that of the genuine coins.

They would still be counterfeit.

I wrote a tongue-in-cheek a while back, after feeling it had been abused

by several posters, almost to the extent that it was not given proper status as a "genuine" number.

OP says only that Delta be any real number between 5 and -5.

Zero is a real number; OP says we should deal with it.

[Guest]

Lovely.

You have a trivial typo "I will take 2 1 coins from piles 11 and 12."

And I'll happily make the change at your request.

- bn

Thanks bonanova. If you'd be so kind as to edit my original post that contains the solution, I'd be most appreciative. I don't want someone to look at it and get very confused. I guess guys with 3k+ post have that kind of power. Thanks

[Guest]

Wouldn't it be the case that if Delta=0, the counterfeit coins cannot be identified?

OP states there are counterfeits; their weight is within limits that include the weight of genuine coins.

Although it could be determined that Delta = 0; weighing would seem powerless to identify the counterfeit pile.

That is, the counterfeits could be struck from material of density equal to that of the genuine coins.

They would still be counterfeit.

I wrote a tongue-in-cheek a while back, after feeling it had been abused

by several posters, almost to the extent that it was not given proper status as a "genuine" number.

OP says only that Delta be any real number between 5 and -5.

Zero is a real number; OP says we should deal with it.

Now that you mention the "different material" issue, I have to agree. Feel free to change the problem to include a "Delta can't equal 0 provision."