[Guest]

A positive integer has the form 12345678901234567890……1234567890 and contains 4000 digits.

At the outset, all digits in odd numbered places starting at the leftmost place are deleted.

Next, all the digits in the odd numbered places in the remaining 2000 digits are removed.

The same operation is continued until no digits remain.

Determine the digit that is the last to be removed.

[Guest]

pretty simple!!!!!!!you will get a pattern in each turn!!!!!!

[Guest]

it is just a bit of iteration, being careful of when there are incomplete cycles of digits. After 11 cycles of elimination, I got the answer below.

........... 8.

Is there a rule that can be evised easily for this?

[Guest]

it is just a bit of iteration, being careful of when there are incomplete cycles of digits. After 11 cycles of elimination, I got the answer below.

........... 8.

Is there a rule that can be devised easily for this?

The original position of the first digit after each iteration is the next highest power of 2, so 2^11 = 2048

You have to be careful about deleting all odd position digits otherwise you wouldn't stop at the 11th iteration.

[Guest]

digits left Original position of the remaining numbers

4000 1 2 3 4 …

2000 2 4 6 8 …

1000 4 8 12 16 …

500 8 16 24 32 …

250 16 32 48 64 …

125 32 64 96 128 …

62 64 128 192 256 …

31 128 256 384 512 …

15 256 512 768 1024 …

7 512 1024 1536 2048 …

3 1024 2048 3072

1 2048

Digit originally in position 2048 was '8'

In Excel terms, the digits left = int(n/2)

The original position is the 2nd & 4th from the row above, and then continue the arithmetic series from there...

[Guest]

General rule :

Let the number be : 1234567890123.... with "n" digits. Here it is a repetative pattern of m=10 digits.

The general answer to this puzzle is 2^k (mod m) such that n >= 2^k > n/2 (>= is greater or equal).

For this particular puzzle n=4000 and m=10. Then value of k such that 4000 >= 2^k > 4000/2 is k=11 , 2^k =2048 and answer is 2048(mod 10) = 8.

[Guest]

Answer is 8.

I arrived as follows

4000 digits 1234578901234567890...

2000 digits 24680 24680 ..(Name it as series 1)

1000 digits 48260 48260 (Series 2)

500 digits 86420 86420... (Series 3)

250 digits 62840 62840 .. (series 4)

125 digits - series 1

62 digits = series 2

31 digits = series 3

15 digits = series 4

7 digits = series 1

3 digits = 482

1 digit = 8

[Guest]

8.

2X2X2X2X2X2X2X2X2X2X2=2048

2048= 204X10 + 8

[Guest]

The last number left corresponds to the highest power of 2 so that would be 2048 so the last # is 8

[Guest]

6

[Guest]

I just did a single series 1-0 and got 8. I added a second and got 8, a third I got 8... Eventually, whenever working it down you will get a repeating series of 4-8-0, which will break down into smaller series of 4-8-0 until you get only 8 or a repeating 8. The answer is 8.

[Guest]

It is not 8. He specified the odd digits from the left should be removed. The true answer is 4.

there are 10 digits in 1234567890, you remove 1/2 : 24680 is what is left to repeat in 2000 digits. From the the left, odd digits are removed: 48 repeating is left for 2/5 of the digits as there were only 5 digits left. now it is 800 digits left, and if you remove odd numbered digits from the left again you get 4 repeating for 400 digits. all you end up with in the end is 4.

Edited December 11, 2009 by Artem1620

[Guest]

It is not 8. He specified the odd digits from the left should be removed. The true answer is 4.

there are 10 digits in 1234567890, you remove 1/2 : 24680 is what is left to repeat in 2000 digits. From the the left, odd digits are removed: 48 repeating is left for 2/5 of the digits as there were only 5 digits left. now it is 800 digits left, and if you remove odd numbered digits from the left again you get 4 repeating for 400 digits. all you end up with in the end is 4.

Except that it is NOT 48 repeating as 24680 is an odd number of digits, so before you continue you have to complete it to an even number to divide in half. 48 doesn't repeat, but 48260 repeats. (2468024680)=48260).

8

[Guest]

I did it all out and got 2

[Guest]

12345678901234567890

2468024680

4848

88

8

simple

no math needed

[Guest]

I'm a math geek, and this takes no math whatsoever

12345678901234567890

2468024680

4848

88

8

simple

no math needed

[Guest]

I wrote some MAtlab code and came up with the answer.

i=1:4000;

while(length(i)>1)

for j=1:length(i)

if(mod(j,2)>0)

i(j)=0;

end

end

n=1;

for k=1:length(i)

if i(k)>0

m(n)=i(k);

n=n+1;

end

end

clear i;

i=m;

clear m;

end

i = 2048 <---answer

[Guest]

Sorry that was the position of the answer, not the answer.

i(1:9)=1:9;

for(c=10:10:4000)

i(c:c+9)=0:9;

end

while(length(i)>1)

for j=1:length(i)

if(mod(j,2)>0)

i(j)=0;

end

end

n=1;

for k=1:length(i)

if i(k)>0

m(n)=i(k);

n=n+1;

end

end

clear i;

i=m;

clear m;

end

i = 8 <--answer

Edited December 12, 2009 by seg9585

[Guest]

Digits:

4000 : 1 2 3 4 5 6 7 8 9 0 . . .

2000 : 2 4 6 8 . . .

1000 : 4 8 2 6 . . .

0500 : 8 6 4 2 . . .

0250 : 6 2 8 4 . . .

0125 : 2 4 6 8 . . .

0062 : 4 8 2 6 . . .

0031 : 8 6 4 2 . . .

0015 : 6 2 8 4 . . .

0007 : 2 4 6 8 . . .

0003 : 4 8 2

0001 : 8 <--- Last digit to be deleted

[Guest]

You guys are all making this way too difficult. All you need to do, is just take the first 10 digits. 1234567890

1. 1234567890

2. 24680

3. 48

4. 8

Easy as that. Because you know automatically that the rest of the digits are going to follow the exact same pattern.

Keep in mind that sometimes, things aren't as hard as they look.

[Guest]

You guys are all making this way too difficult. All you need to do, is just take the first 10 digits. 1234567890

1. 1234567890

2. 24680

3. 48

4. 8

Easy as that. Because you know automatically that the rest of the digits are going to follow the exact same pattern.

Keep in mind that sometimes, things aren't as hard as they look.

So, regardless of the length of the initial integer ( > 9 ), the last digit to be removed will always be 8?