mvz2102 wrote:A question came up in the OG last night that required you to get the nth decimal of some arbitrary fraction, I believe it was 6/11.

I got the correct answer, but my method was long and difficult and involved some sketchy guesswork. Is there any way to do this for any general fraction of two integers a/b?

If that type of question appears on the GMAT, it will always involve an infinitely repeating decimal; you need to find the pattern (usually through long division), then apply the pattern to the question asked.

Let's say the question is: "What's the 113th digit to the right of the decimal point in the decimal representation of 6/11?

We think of 6/11 as "6 divided by 11" and use long division to get:

6/11 = 0.545454...

Here we see that it's a 2 digit repeating pattern - odd numbered digits are "5" and even numbered digits are "4". Since we want the 113th digit, an odd numbered digit, so it will be "5".

Here's a trickier one:

"What's the 23rd digit to the right of the decimal point in the decimal representation of 1/7?"

We think of 1/7 as "1 divided by 7" and use long division to get:

1/7 = .142857142....

Here we see that it's a 6 digit repeating pattern. We want the 23rd digit. "23/6" is "3 and 5/6"; the 5/6 tells us that we want the 5th digit in the pattern, which is "5".

edited to change "left" to "right"!

Thanks Stuart. I realized it was a repeating decimal, and that I'd be able to get the answer if I could just expand it... but I wasn't sure how. Since you mentioned long division, a little refresher in 6th grade math got me up to speed.

While you're around, there are 2 DS questions in my OG that I can't figure out for the life of me:

142. If x is an integer, is 9^x + 9^-x = b?

i) 3^x + 3^-x = sqrt(b+2)
ii) x > 0 = b

Statement one is clearly sufficient. My interpretation of statement two is "x is greater than 0, and 0 is equal to b". If this is accurate, then the right side of the equation is zero, and since 9^x + 9^-x will always be greater than zero, this allows you to definitively answer NO. If so, then the answer should be E. The OG however, says B is not sufficient, therefore making the answer A.

146. Is n an integer?

i) n^2 is an integer
ii) sqrt(n) is an integer

They should both be sufficient to answer YES. This seems incredibly simple. OG answer says A.

this is example of repeting sequence.u just need to check after how many digits sequence is repeting.ex
ex 105th digit of 5/11=.45454545.... so divide 105 by 2 remainder is 1.so first digit 4 is ur answer.

mvz2102 wrote: While you're around, there are 2 DS questions in my OG that I can't figure out for the life of me:

142. If x is an integer, is 9^x + 9^-x = b?

i) 3^x + 3^-x = sqrt(b+2)
ii) x > 0 = b

Statement one is clearly sufficient. My interpretation of statement two is "x is greater than 0, and 0 is equal to b". If this is accurate, then the right side of the equation is zero, and since 9^x + 9^-x will always be greater than zero, this allows you to definitively answer NO. If so, then the answer should be E. The OG however, says B is not sufficient, therefore making the answer A.

146. Is n an integer?

i) n^2 is an integer
ii) sqrt(n) is an integer

They should both be sufficient to answer YES. This seems incredibly simple. OG answer says A.

For others, those are questions 165 and 167 in the 12th edition of the OG.

For your q146, the answer in the OG is B, not A (perhaps the old edition had a typo). Statement (1) isn't sufficient because all it says is that n is the root of an integer, which may or may not be an integer.

For example, if n^2 = 4, then n =2, which is an integer; if n^2 = 3, then n=root3, which isn't an integer.

For your q142, statement (2) is simply "x > 0" in the 12th edition.