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I am going to solve this in not so smart way is what I presume because I happen to know 27*37=999

and anything of the form as below:

x/9 = 0.xxxxxxxx....

xy/99= 0.xyxyxyxyxyxy.......

xyz/999 = 0.xyzxyzxyzxyzxyz.......


and so 5*(37)/27*(37) = 185/999
and so 185/999 =0.185185185185.....

from here very easy to say that 100th digit to the right of the decimal point = 1


I feel it is not the smartest way since I already knew 27*37 which is 999 and hence you now know what I did there.
Anyways, this works fine.

Answer is Option (B)
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Nice solutions above. Just wanted to add a quick general framework for repeating decimal questions since the GMAT Focus loves this pattern.

The recipe is always the same three steps: (1) find the repeating block by doing the long division, (2) find the length of that block, and (3) use modular arithmetic to pinpoint your digit.

For 5/27, the repeating block is "185" with length 3. To find the 100th digit, divide 100 by 3: you get 33 remainder 1. That remainder tells you it's the 1st digit of the repeating block, which is 1.

Quick shortcut to remember: if the remainder is 0, it's the LAST digit of the block (not the first). That trips up a lot of people. For example, the 99th digit here would be remainder 0, meaning it's the 3rd digit = 5.

Also worth noting — fractions with denominators that are factors of 9, 99, 999, etc. always produce clean repeating decimals. Recognizing 27 as a factor of 999 (since 27 x 37 = 999) can speed things up.

Answer: (B) 1
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aditya1818
I am going to solve this in not so smart way is what I presume because I happen to know 27*37=999


I'm not seeing what makes this any less than smart. You used a[n uncommon] memorized fraction-decimal conversion, but your solution to the problem isn't just rote recitation; you still have to perform the same pattern-recognition task as everybody else does.

While GMAC's official problems will never require obscure memorized knowledge, there's nothing un-smart about, or otherwise wrong with, taking advantage of such knowledge if you happen to have it and are served a problem to which it's relevant.

(Just make sure you aren't trying to memorize ungodly amounts of information in the vague hope that you might be able to put some of it to use on the official test. Most GMAT problems are completely memorization-proof—and the small handful to which you might be able to apply memorized knowledge won't really reward you for it anyways. On this problem for instance, manipulating the pre-existing fact that 27 x 37 = 999 doesn't save any significant time or work relative to just grinding out the long division until the pattern pops out.)
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I mean I am a sucker for things like this - I automatically memorize too! 😂
It is so awesome it is 999

Kudos aditya1818


RonPurewal



I'm not seeing what makes this any less than smart. You used a[n uncommon] memorized fraction-decimal conversion, but your solution to the problem isn't just rote recitation; you still have to perform the same pattern-recognition task as everybody else does.

While GMAC's official problems will never require obscure memorized knowledge, there's nothing un-smart about, or otherwise wrong with, taking advantage of such knowledge if you happen to have it and are served a problem to which it's relevant.

(Just make sure you aren't trying to memorize ungodly amounts of information in the vague hope that you might be able to put some of it to use on the official test. Most GMAT problems are completely memorization-proof—and the small handful to which you might be able to apply memorized knowledge won't really reward you for it anyways. On this problem for instance, manipulating the pre-existing fact that 27 x 37 = 999 doesn't save any significant time or work relative to just grinding out the long division until the pattern pops out.)
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Caught in the trap of 100th place digit vs 100th digit

RonPurewal
In the decimal expansion of the fraction 5/27, what is the 100th digit to the right of the decimal point?

A. 0
B. 1
C. 3
D. 5
E. 8

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To find the 100th digit to the right of the decimal point in 27/5, we first expand the fraction.

27/5=0.185185185...

Notice that the digits 185 repeat continuously.

So, the repeating block is:

185

This means:

1st digit = 1
2nd digit = 8
3rd digit = 5
4th digit = 1
5th digit = 8
6th digit = 5
and so on.

The cycle length is 3.

Now 99th digit will be=5

Therefore, the 100th digit is= 1


Possibility of Mistakes
1. Not going beyond 2 or 3 divisions
Students may stop early and fail to recognize the repeating pattern 185.

2. Not reading carefully: “100th place digit” vs “100th digit”
“100th digit” means the digit in the 100th position after the decimal, not the “hundredths place” digit.

3. Incorrectly concluding the 100th digit is 5 or 1 without using remainders properly
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