# Manhattan GMAT Challenge Problem of the Week – 1 June 2010

Here’s the latest Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

**Question**

The three-digit positive integer

ncan be written asABC, in whichA,B, andCstand for the unknown digits ofn. What is the remainder whennis divided by 37?

(1)

A+B/10 +C/100 =B+C/10 +A/100(2)

A+B/10 +C/100 =C+A/10 +B/100a. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

b. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

d. EACH statement ALONE is sufficient.

e. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

**Answer**

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 × *A*.

Now, 111 factors into 3 × 37, so *n* = 3 × 37 × *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.

**The correct answer is (D): EACH statement ALONE is sufficient.**

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## 10 comments

Govardhan on June 1st, 2010 at 9:58 pm

@Clay,

In RHS of st 1, (B + C/10 + A/100), how do we know that B stands for hundredth position or C for tenth position or A for unit's place?

As per GMAC syllabus, what topic does this question cover?

raghavendra on June 2nd, 2010 at 1:57 am

A, B and C are supposed to be integers from 0-9 . that is single digits, since n is a 3 digit integer.

Thus,(B + C/10 + A/100) means b in units place, c in first decimal place and a in 2nd decimal place.

and no idea wat part of the gmac syllabus it covers .

Stacey Koprince on June 3rd, 2010 at 8:47 am

raghavendra is correct in his explanation of the place values for these numbers. If you're not sure about the theory, try a few real numbers to see how it works out.

Let's take the number 456, where A=4, B=5, and C=6. For (B + C/10 + A/100), plug in the numbers: 5 + 6/10 + 4/100 = 5.64.

In future, do know that taking a digit and placing it as the numerator over the number 10 as the denominator is one way of representing the tenths place of a number. Similarly, taking a digit and multiplying it by 10 is one way of representing the tens place (not tenths).

This kind of question covers Number Properties. It is testing us on our knowledge of place value, divisibility, and remainders.

Jules Y on June 3rd, 2010 at 8:48 pm

Hi Caitlin, I have a question. How does the 2 statements guarantee that n must be equal to 111 and not 222 or 333? Thanks.

Stacey Koprince on June 4th, 2010 at 6:59 am

The solution does not state that n = 111. It states that n = 111*A. That is, whatever A is, n equals A times 111. So n itself is either 111 or some multiple of 111 (possibly 222, or 333, or so on).

It turns out that 37 is a factor of 111. If 37 is a factor of 111, then 37 is also a factor of any multiple of 111. Therefore, 37 is a factor of 111*A, which equals n. Therefore, 37 is a factor of n.

37 will always divide evenly into whatever n is because 37 is a factor of n.

abhijeet on June 12th, 2010 at 7:51 am

this is the very first problem i tried to solve, and i got the correct answer, I m happy..

>Jules Y on June 3rd, 2010 at 8:48 pm

>

>Hi Caitlin, I have a question. How does the 2 statements guarantee >that n must be equal to 111 and not 222 or 333? Thanks.

the statement doesnt state anything related to the equality of A B and C. but solving statements we can conclude that, ABC=BCA, which shows that A=B, B=C and C=A.. thus all the digits are equal. So the value turns to be 111,222,333. etc.

Rochelle Ballard on June 15th, 2010 at 6:49 am

I'm not sure why you have to actually come up with a formula? If the DS quesiton is asking - whether or not you have enough informaiton to find out the remainder when n is divided by 37? all you need to find out is if you have enough information to derive a "3 digit integer".

I stopped trying to solve the problem after I realized all the digits had to be the same. I didn't think about A having to be greater than 0. Is this incorrect?

Stacey Koprince on June 15th, 2010 at 7:52 am

I'm not sure which formula you're referring to, so I'm not entirely sure that I understand what you're asking. The only formula in the explanation after the point where we realize the digits are the same is "n = 3 × 37 × A" - and that's not a formula we're trying to solve or manipulate. We're just writing out a representation for n.

The key to the above solution was being able to say that some part of what makes up n is divisible by 37, which is why we wrote out that representation of n. If we know that some part of what makes up n is divisible by 37, then we know that n itself is always divisible by 37 (making the remainder always zero when dividing by 37), regardless of what n actually is.

If you are talking about another formula, can you please clarify your question?

Also, you mention "all you need to find out is if you have enough information to derive a "3 digit integer". Again, I'm not quite sure I fully understand what you're saying here. Were you checking to see whether we have enough information to derive ONE 3-digit integer?

We don't have enough info to do that; we only have enough info to derive a representation that narrows down the possible values for n, but does not give us only one possible value. It just happened that all of the possible values returned the same answer to the original question. What if the information had allowed us to say that n was either 111 or 150?

Re: the A can't be zero issue, on this one, we would have gotten away with not noticing that A couldn't be zero, yes. If you realize all of the digits are the same, then n could be 0, 111, 222, and so on. 0 is also divisible by 37, just as all of the other possible values for n are, so the answer would have stayed the same. On another problem, though, having an extra possible value in the mix might have changed things, so it's still a good idea to explore what the problem allows you to deduce.

Rochelle Ballard on June 15th, 2010 at 8:24 am

ok that makes sense now actually. my last question is how is one to know that you should factor out 111?

Stacey Koprince on June 15th, 2010 at 8:46 am

That question can be split into two parts, actually. So, I get to the point that I know I can represent n as AAA, where each A represents the same single digit from 1 to 9. That gives me a range of possible values for n: 111, 222, 333, 444, 555, 666, 777, 888, 999.

Now, somehow, I have to test what the remainder is for each of those numbers when divisible by 37 - I want to know whether 37 is a factor for all (in which case, my answer is "sufficient" because the remainder is always zero) or whether 37 is a factor for some but not all (in which case, my answer is "not sufficient" because sometimes the remainder is zero and sometimes it is something else) or whether 37 is never a factor (in which case, I have more work to do, because I don't yet know whether I always have the same remainder or whether it is sometimes different).

I *really* don't want to do long division 9 times - ugh! But I do notice a pattern - maybe there's a shortcut! Maybe I can use the pattern to figure out whether 37 is a factor of some, all, or none of these possible values of n. What's the pattern?

111*1

111*2

111*3

111*4

111*5

111*6

111*7

111*8

111*9

So, I don't know which digit I'm ultimately multiplying by, but I do know that 111 is in the mix for every single possible value of n. So then I ask myself: does that tell me anything? What are the factors of 111?

Hmm. 111 factors to... 3*37. Bingo!

So, I wouldn't necessarily know for sure that it would help me when I first start going down the above path. But I do know that there has got to be an easier way to solve the problem than doing long division 9 times - that's not reasonable in 2 min. And patterns are a very typical way to save time on this test. So I pursue the pattern to see whether it helps and... voila, it does!