Problem Solving

Hi Vincen,

This question is built around the concept of 'remainders.' For example, 7/2 = 3 remainder 1. In this prompt, we're told that 2 calculations would have the SAME remainder:

P/Q
Q/P

The 'easiest' remainder is 0. That occurs when the denominator divides evenly into the numerator. For BOTH of those calculations to have a remainder of 0, the two values would need to be EQUAL to one another... meaning that (P)(Q) would be a PERFECT SQUARE. There's only one answer that fits...

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Vincen wrote:If p and q are positive integers, and the remainder obtained when p is divided by q is the same as the remainder obtained when q is divided by p, which of the following is a possible value of pq?

(A) 62
(B) 55
(C) 42
(D) 36
(E) 24

The remainder obtained when p is divided by q is the same as the remainder obtained when q is divided by p
This information is indirectly telling us that p = q
To explain why, let's see what happens if p does NOT equal q
If that's the case, then one value must be greater than the other value.
Let's see what happens IF it were the case that p < q.

What is the remainder when p is divided by q?
Since p < q, then p divided by q equals 0 with remainder p

IMPORTANT RULE: When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0

What is the remainder when q is divided by p?
Based on the above rule, we know that the remainder must be a number such that 0 ≤ remainder < p

Hmmmmm. In our first calculation (p ÷ q), we found that the remainder = p
In our second calculation (q ÷ p), we found that 0 ≤ remainder < p
Since it's IMPOSSIBLE for the remainder to both EQUAL p and BE LESS THAN p, we can conclude that it's impossible for p to be less than q.

Using similar logic, we can see that it's also impossible for q to be less than p.

So, it MUST be the case that p = q
So, pq = p² = the square of some integer

Check the answer choices . . . only D is the square of an integer.

Vincen wrote:If p and q are positive integers, and the remainder obtained when p is divided by q is the same as the remainder obtained when q is divided by p, which of the following is a possible value of pq?

(A) 62
(B) 55
(C) 42
(D) 36
(E) 24

Since we've already seen in the above posts that pq COULD equal 36, let's find some actual values that satisfy the given information.
Notice that, if p = 6 and q = 6, then p divided by q leaves remainder 0, AND q divided by p also leaves remainder 0
Here, pq = (6)(6) = 36

Cheers,
Brent

Vincen wrote: ↑

Sun Oct 08, 2017 11:13 am

If p and q are positive integers, and the remainder obtained when p is divided by q is the same as the remainder obtained when q is divided by p, which of the following is a possible value of pq?

(A) 62
(B) 55
(C) 42
(D) 36
(E) 24

The OA is D.

I got confused here. I need an explanation. Experts, help.

If p = 6 and q = 6, the remainder is the same for p/q and q/p.

Answer: D