[Am I wrong…2]

16 - 36 = 25 - 45

4^2 - (9x4) = 5^2 -(9x5)

4^2 - (9x4) + 81/4 = 5^2 -(9x5) + 81/4 ......by adding 81/4 to both sides

(4-9/2)^2 = ( 5-9/2)^2......we can cancel the powers on both sides

4- 9/2 = 5- 9/2............add 9/2 to both sides to get:

4=5 !!!

Am I wrong?

[Am I wrong?]

Let X= 0.9999.............

then 10X=9.99999..........

10X- X = 9.99999......... - 0.999999.......

9X = 9

that means:

X=1

so

1= 0.999999........

the same is in case of:

1/3+1/3+1/3 =3/3 = 1

but, 1/3 = 0.33333.........

so

0.333333......+0.33333.....+0.33333.......= 0.999999.......

which is < 1

Am I wrong?

[What is it?]

church

right..thx

[What is it?]

something has:

ch to right,ch to the left and you are in the middle!

what is it?

[The missing number]

131

Call the sequence S. Then S(1)=10, S(2)=31,...

It appears that S(n) is the sum of the (3n)th prime, the (3n-1)th prime, and the (3n-2)th prime. So, S(5) is the sum of the 15th prime, the 14th prime, and the 13th prime (47+43+41), which is 131.

Thats it...Very good!

[The missing number]

Try to find the missing number :-

10,31,59,97,.........,173

please give your reason....thanks.

[Happy New Year!]

Happy new year, Wolfie!

God bless and keep you.

Thank you dear.....

[Happy New Year!]

I wish you all, a very happy new year,may all your wishes come true...

God bless you...

welcome 2012.... ready...

3.....2.....1......Gooooooooo!!

[Merry Christmas!]

Meeeeeeeeeerrrrrrrrry Christmas everyone.

Glory to God in the highest, and on earth peace, good will toward men.

Jesus love you ALL

ESS for 553

write 6L on a piece of paper, turn upside down and it'll look like 79. all other figure are 6L +79

8SI= 158

LE2= 237

etc

Very good!

ESS

A snake sound!

You are FANTASTIC!!!!!

Well...it seems to be easy.....or it may be hard if you don`t sit at the other side....lol

What is next to these combinations?

6L,8SI,LE2,9IE,S6E,hLh,.........

note: you need a little bit fantasy...

Using the grid mentioned by CaptainEd, I entered numbers 1-16 in the grid.

4 14 15 1

9 7 6 12

5 11 10 8

16 2 3 13

And I find that each row, column and diagonal sum up to 34

Thats right...and you can enter numbers in the grid according to my combination also.

So I would be allowed to keep placing marbles on the two balances until they both said 49g?

Once I have done that I have 100g. The actual weights would be 48g and 52g.

Using your sample weights:

-3   +1

5.4  5.9

5.6  6.1

6.3  6.3

7.6  6.8

8.2  7.1

9.2  7.5

9.7  8.3

Say I pick the (-3) balance first. Placing the first set of marbles will make is say 49g. I then place the rest on the second balance for another 49g.

Thats right!....but according to my method ,I was to weigh 102 and 96gms

Then using the scale balance I was to devide the total into two equal parts....each one would be 100 gm.

.................. +1........... -3

102............101..........105

96...............95............99

(101+99)/2=100

(105+95)/2=100

Thanks

If I apply both curr and scottbarnes logic I would only need to use one balance

put the smallest marble on one of them, it well weigh either more or less than 5 (if we assume that the smallest is less than 8g) if more then I will continue adding marbles till I reach 101, if less, then I'll add till 97

which is why we can't weigh a specific mass of marbles, otherwise it would be too easy.

@wolfgang, I can easily tell which balance is which

If there are no weights on the digital balances the should give readings of 1 and -3, it will only show 0 at 0 if there is a ratio error i.e weighing double, but this is an addition error, and I don't think that taking the zero readings is considered using the balances

when I said (more than 5 gms) that may be true if the smallest one is over 10gm either,so we can not estimate the weight and we shouldn`t.

In case of No weights of the digital balances, both of them are calibrated to read (0).

Keep placing marbles until each digital scale says 49g? Or is that considered more than once use for a scale just trying to get to 49g?

I was making an assumption that given a large sample of varying marbles some trial would be required to find marbles that can be combined to total 100g. If that is not allowed can you expand upon the problem? Because with the wording given it could be impossible to exactly measure 100g.

Say there are 20 marbles each weighing 5.5g. It would be impossible to get 100g.

It is possible to make a set of marbles weigh different amounts above 5g that can never be combined in such a way to total 100g exactly.

With this puzzle it should be assumed that it is possible to combine some marbles to total 100g. But how else can you find them amongst those that can't total 100g.

Keep placing marbles on a given balance till a given weight is to be concidered as one weight.

I´ll give you an example of 14 marbles having a total weight of 100 gm:-

5.6gm,6.3gm,7.1gm,6.8gm,8.2gm,5.9gm,8.3gm,7.5gm,9.7gm,9.2gm,6.3gm,7.6gm,6.1gm,and 5.4gm

He thinks you gave the instructions to a magic square.

1d,4b,4c,1a,3a,2c,2b,3d,2a,3c,3b,2d,4d,1b,1c,4a

  4 14 15  1

  9  7  6 12

  5 11 10  8

16  2  3 13

Which your new hint supports.

Yes...I gave the instructions to a magic square,by giving the squares on X-axes,letters(a,b,c,and d)

and giving numbers(1 to 4) to Y-axes squares.

now,1d will mean the square for 1,

and 4b will indicate where number 2 should be,,,and so on

Not sure why you would need the scale balance since it can be done with just the two digital.

Let x be the displayed weight and y be the actual weight.

d1: x1 = y1 + 1

d2: x2 = y2 - 3

Since we do not know which is d1 and which is d2: x1 = x2

Since the goal is 100g: y1 + y2 = 100 => y1 = 100 - y2

y1 + 1 = y2 - 3

100 - y2 + 1 = y2 - 3

2 * y2 = 104

y2 = 52

y1 = 48

x1 = x2 = 49

Measure out 49g on each digital balance. One will actually be 48g and the other will be 52g for a net of 100g.

I suppose you can use scale balance to determine which digital is which based upon which set is heavier.

Sorry...but how can you measure the 49gm?how can you say that this amount of marbles are 49gm?

we can obtain ten times 34 out of this sequence .

Sum all the series from 1a to 4d, and divide them to 8, there is your 34

?? ..what??

Its the old question of:

"If I asked the other guard which door leads to freedom, what would he say?"

Only need to ask one guard

Always take the other door.

I agree with "The Rogue"

if you were to weigh a single (smallest) marble first, it's actual weight a little over 5gm, we would get a reading of either 6 (if scales are +1) or 2 (if scales are -3). should we get a read of 2 we would know that this is the -3 scales as we know the minimum weight of a marble is 5. that just leaves us to weigh out to 101 on the other scales (or vice versa) leaving the balance scales completely alone.</p>

Nice try...but you should use all of them....ty

First take any one balance and put marbles on it till it reads 100.

Case 1: You chose the accurate one and the actual wight is also 100.

Case 2: You chose the +1 one and the actual weight is 99.

Case 3: You chose the -3 one and the actual weight is 103.

Then choose another balance and weigh the same set of marbles.

Case 1: Your first balance was accurate and actual weight is 100.

Hence your second choice is either +1 or -3.

So it will display either 101 or 97.

Case 2: First balance was +1 and actual wight is 99.

Your second one is either accurate or -3.

So it will display either 99 or 96.

Case 3: First balance was -3 and actual weight is 103.

Second choice is either accurate or +1.

So it will display either 103 or 104.

From the reading on second balance, we can figure out which balance was used first and second.

Then put the same set of marbles on the last balance.

If the last balance is accurate put on/remove marbles till it becomes 100.

If it is +1, adjust so that it reads 101.

If it is -3, adjust so that is reads 97.

Dear SMV!....only 2 balances are digital,the third one is a scale balance with two arms.

If you were asked to weigh exactly 100gm of different sized Marbles(each one of them is more than 5gm), using three balances each one only once,these balances are as follows:

one ordinary scale balance,which is accurate.

and two analytical balances,each one of them has an error,the first one displays (one)gm more..i.e.,if you put 100gm on it, it will show 101gm.,while the other balance displays (three) gm less,...i.e.,by putting 100 on it ,it will display 97gm.and you don`t know which is which..

How can you weigh 100gm...?

I sure don't FEEL close!

By the way, there's a LOT more symmetry--the sequence appears to be a Hamiltonian circuit around a 2x2x2x2 hypercube. But I'm still missing a couple of fundamentals.

1) Perhaps we are to add one or two dimensions (egad, maybe 4 more dimensions!)

2) But his sequence always moves to a new dimension before finishing the old ones. However, this sequence has now completely filled the 2x2x2x2 without starting into new ones.

3) and I have no idea how 1-4 relate to a-d. If we see them as hexadecimal numbers, the gap between 4 and A is not a convenient one, I could be more comfortable (still clueless, but more comfortable) if a-d had been c-f.

This is very interesting wolfgang, thanks for the workout...still working...

Thank you Dear...I am so glad to hear this....sometimes,the situation is not so complex as we immagine!...take it easier....!!!