Remainders with Variables for the GMAT, GRE, SAT

A refresher on remainders follows the solution.

When integer b is divided by 13, the remainder is 6. Which of the following cannot be an integer?
A) 13b/52
B) b/26
C) b/17
D) b/12
E) b/6

“When b is divided by 13, if the remainder is 6”
Let’s take that to mean the following:
b/13 = q remainder 6
q will be an integer since it’s essentially the answer to the equation.

Rearrange, remembering that remainders stay the same number when you multiply by the divisor (which is 13 in this case)
b = 13q + 6

Plug in the answers so we can see which works best.

A) 13b/12

Will 13(13q + 6)/12 equal an integer if we plus in for q?

Let’s say q = 6.

If q = 6, then 13q + 6 = 84, which is is divisible by 12; hence 13(84)/12 is an integer.

B) b/26

Could (13q + 6)/26 result in an integer for an integer value of q?

First off, 26 is exactly 2 times 13. So, for any integer value of q, 13q will either be divisible by 26 if q is even or produce a remainder of 13 if q is odd.

Adding 6 to 13q, the expression will either produce a remainder of 6 or 19, but will never produce a remainder of zero. So, (13q + 6)/26 = b/26 can never equal an integer.

B.

Refresher Question

P is a two-digit number with exactly two factors. What is the remainder when (P² — 1) is divided by 6?

A) 0
B) 2
C) 4
D) 6
E) Cannot be solved.

P is a two-digit number, with exactly two factors.

Any number with exactly two factors is a prime number.
So P is a two-digit prime number.
Also, P is an odd number (which we should know since any prime number greater than 2 is an odd number).

Now for the remainder, when P² — 1 is divided by 6.

Let’s use the general property of any prime number that is greater than 3:

We can express that number in 6k + 1 or 6k — 1 form, where k is a positive integer, similar to how we used variables in the original question above.

As P is a two-digit prime number, we can also express P in the form 6k + 1 or 6k — 1.

Example 1: P = 6k +1,
P² = (6k + 1)² = 36k² + 12k + 1

Or, P² — 1 = 36k² + 12k

Or, P² — 1 = 6 (6k² + 2k)

P² — 1 is a multiple of 6.

Iif we divide P² — 1 by 6, the remainder will be 0.

Example 2: P = 6k — 1
P² = (6k — 1)² = 36k² — 12k + 1

Or, P² — 1= 36k² — 12k

Or, P² — 1 = 6 (6k² — 2k)

We can see that P² — 1s a multiple of 6. Hence, if we divide P² — 1by 6, the remainder will be 0.

Therefore, we can see that whether P is 6k + 1 or 6k — 1, the remainder when P² — 1 is divided by 6 is always 0.

A.