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# S96-06

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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:50
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9
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Difficulty:

85% (hard)

Question Stats:

47% (01:38) correct 53% (02:09) wrong based on 116 sessions

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The three-digit positive integer $$n$$ can be written as $$ABC$$, in which $$A$$, $$B$$, and $$C$$ stand for the unknown digits of $$n$$. What is the remainder when $$n$$ is divided by 37?

(1) $$A + \frac{B}{10} + \frac{C}{100} = B + \frac{C}{10} + \frac{A}{100}$$

(2) $$A + \frac{B}{10} + \frac{C}{100} = C + \frac{A}{10} + \frac{B}{100}$$

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Joined: 02 Sep 2009
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16 Sep 2014, 01:50
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1
Official Solution:

The question stem tells us that the positive integer $$n$$ has three unknown digits: $$A$$, $$B$$, and $$C$$, in that order. In other words, $$n$$ can be written as $$ABC$$. Note that in this context, $$ABC$$ does not represent the product of the variables $$A$$, $$B$$, and $$C$$, but rather a three-digit integer with unknown digit values. It is important to note that since $$A$$, $$B$$, and $$C$$ stand for digits, their values are restricted to the ten digits 0 through 9. Moreover, $$A$$ cannot equal 0, since we know that $$n$$ is a "three-digit" integer and therefore must be at least 100.

We are asked for the remainder after $$n$$ is divided by 37. We could rephrase this question in a variety of ways, but none of them are particularly better than simply leaving the question as is.

Statement (1): SUFFICIENT. We can translate this statement to a decimal representation, which will be easier to understand. The left side of the equation, in words, is "$$A$$ units plus $$B$$ tenths plus $$C$$ hundredths." We can write this in shorthand: $$A.BC$$ (that is, "A point BC"). After performing the same translation to the right side of the equation, we can see that we get the following:
$$A.BC = B.CA$$

Since $$A$$, $$B$$, and $$C$$ stand for digits, we can match up the decimal representations and observe that $$A = B$$ and $$B = C$$. Thus, all the digits are the same.

This means that we can write $$n$$ as $$AAA$$, which is simply $$111 \times A$$.

Now, 111 factors into $$3 \times 37$$, so $$n = 3 \times 37 \times A$$. Thus, $$n$$ is a multiple of 37, and the remainder after division by 37 is zero.

Statement (2): SUFFICIENT. Again, we can match up the decimal representations of the given equation and find that all the digits are the same. The logic from that point forward is identical to that shown above.

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Joined: 16 Apr 2015
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11 May 2015, 18:39
So, if my understanding is correct, it does not matter if A, B, C is 111, 222, 333, 444, etc. because all of those are contain the prime numbers, 3x37 - correct?
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Joined: 16 Apr 2015
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16 Jun 2015, 09:00
I really liked that question. Very well written and explained.
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Joined: 12 Nov 2016
Posts: 139
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33

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12 Mar 2017, 19:00
Why I am impressed by 37 - what a magic number! 111, 222, 333, 444, 555, 666, 777, 888, 999 - all divided w/out remainder!
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Joined: 30 Jun 2017
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15 Aug 2017, 11:35
This is a great question. However, while attempting the test I did not get the air of what is happening in this particular problem.
Lessons learnt -
1) 37 divides 111, 222, 333, 444,...,999 completely.
2) The way to represent a three digit number having all digits equal (premise of the question).
Intern
Joined: 16 Jul 2013
Posts: 33
Location: Hungary
Concentration: Entrepreneurship, Marketing
GMAT 1: 720 Q48 V40
GPA: 3.37

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05 Dec 2017, 14:25
I solved it differently. A bit more time consuming, but that's what came to my mind.

From the stem we know that:

ABC = n = 100A + 10B + C

1)
100A + 10B + C = 100B + 10C + A
10B = 11A - C

Plugging this into the stem equation we get:
100A + 11A - C + C = n = 111A

I divided 111 by 37, and voila!

2) Same method gives: n = 111C
Re: S96-06 &nbs [#permalink] 05 Dec 2017, 14:25
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# S96-06

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