BTGmoderatorDC wrote:If (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

A. 8
B. 9
C. 16
D. 23
E. 24

If y > 16, (16 # y) = 16, so we need only check the values from 1 to 15
Also, we need not check the FACTORS of 16, since they will all yield a remainder of 0
We're left with:
(16 # 3) = 1 KEEP!
(16 # 5) = 1 KEEP
(16 # 7) = 2
(16 # 9) = 7
(16 # 10) = 6
(16 # 11) = 5
(16 # 12) = 4
(16 # 13) = 3
(16 # 14) = 2
(16 # 15) = 1 KEEP

3 + 5 + 15 = 23

Answer: D

BTGmoderatorDC wrote:If (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

A. 8
B. 9
C. 16
D. 23
E. 24

OA D

Source: Manhattan Prep

16/15 has a remainder of 1.

16/5 has a remainder of 1.

16/3 has a remainder of 1.

So the sum of all possible values of y is 15 + 5 + 3 = 23.

Alternate Solution:

We are looking for all values of y such that 16 divided by y produces a remainder of 1. Then, 16 - 1 = 15 must be divisible by y. Excluding y = 1 (which produces a remainder of 0); the possibilities for y are 15, 5 and 3. Thus, the sum of all possible values of y is 15 + 5 + 3 = 23.

Answer: D