remainder problem

by **navdeep_bajwa** Wed Sep 30, 2009 7:51 pm

MGMAT Challenge Problems
08/12/02

DS Question
sqrt ABC=504
Is B divisible by 2?

(1) C = 168

(2) A is a perfect square

(A) 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.

Solution

Divisibility problems can be solved using prime factorization.

The prime factorization of 504=2^3*3^2*7

Therefore, using the given equation, we can see that:

sqrt ABC=504

ABC=2^6* 3^4*7

To answer the question, we must determine whether B contains one of the six 2's in the prime factorization.

Statement (1) alone tells us that 168=2^3*3*7

This tells us that C has three of the 2's in the prime factorization. However, since we have no information about A or B, this is not sufficient information to answer the question.

Statement (2) alone tells us that A is a perfect square.

This tells us that if A contains any 2's as prime factors, it must have an even number of 2's. (The only way a number can be a perfect square is if its prime factors come in pairs). This, again is not sufficient information to answer the question.

Using both statements together, we know that C has three of the 2's. We also know that A can have either zero of the 2's or two of the 2's, but, since A is a perfect square, it cannot have all three of the remaining 2's.

Thus, B must have at least one 2 as a prime factor. The correct answer is C.

Why answer is C not E

What about if A is 16 or 25 then B becomes odd number

by **RonPurewal** Thu Oct 01, 2009 7:51 am

Please cite the source (author) of this problem. We cannot reply unless a source is cited (and, if no source is cited, we will have to delete the post!). Thanks.

in brief: if you have those two statements together, then a cannot be 16 or 25, so your examples are inapplicable.

by **navdeep_bajwa** Thu Oct 01, 2009 1:19 pm

hi Sir,

This question is actually from manhattan challenging problems set

by **navdeep_bajwa** Sun Oct 04, 2009 2:28 pm

hi this is from MGMAT challenge problems posted on 8/12/02

Please advice

by **esledge** Wed Dec 30, 2009 5:51 pm

navdeep_bajwa Wrote:Why answer is C not E

What about if A is 16 or 25 then B becomes odd number

Given: ABC = 504^2 = (2^6)(3^4)(7^2)

If C = 168 = (2^3)(3)(7),
then AB = ABC/C = (2^6)(3^4)(7^2)/(2^3)(3)(7) = (2^3)(3^3)(7).

If A = 16, then B = AB/A = (2^3)(3^3)(7)/(2^4) = (3^3)(7)/2 = fraction! Is B even? No.

If A = 25, then B = AB/A = (2^3)(3^3)(7)/(5^2) = fraction! Is B even? No.

Maybe that's what you were getting at? If you assume A, B, and C must be integers, the answer is always "Yes" for the combined statements and therefore the answer is (C), but if you don't assume B is an integer, the answer is (E).

I think this problem was written with the intent that A, B, and C are constrained to be integers (on most even/odd questions, the GMAT will explicitly rule out fractions from consideration). We'll take a closer look at it to make sure we have the right constraint specified. Thanks!

by **StaceyKoprince** Tue May 18, 2010 2:39 pm

then AB = ABC/C = (2^6)(3^4)(7^2)/(2^3)(3)(7) = (2^3)(3^3)(7).

If A = 16, then B = AB/A = (2^3)(3^3)(7)/(2^4) = (3^3)(7)/2 = fraction! Is B even? No.

If A = 25, then B = AB/A = (2^3)(3^3)(7)/(5^2) = fraction! Is B even? No.

I disagree, actually - I think this one's okay.

The first line from Emily's post, above, is fine. It tells us that, given the info given in the question stem and statement 1, A and B together must contain the following primes and only the following primes: (2^3)(3^3)(7).

The next line then says "If A = 16." A can't be 16, because 16 = 2^4. We don't have four 2's available; we only have three 2's available.

The final line says "If A = 25." A can't be 25 either, because 25 = 5*5, and we don't have any 5's available.