If (x # y) represents the remainder that results when the...

[kyma343]

This if from the last CAT that i did.

Here's the question:
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

Does this question mean that the numbers from 16 to y would be added and then see if it equals y? I have a hard time pinpointing what they're asking. When x/y it will yield the sum from 16 to y?

Please help.

Thanks,
KM

[thoppae.saravanan]

The question simply asks you to find the sum of all possible values of y such that (16#y) = 1.

From the first part of the question (16#y) is equal to remainder when 16 is divided by y.

So the possible values of y could be 3,5,15 because dividing 16 by these numbers will give you remainder 1. So, sum of all these values is 23. So option D.

[Ben Ku]

This if from the last CAT that i did.

Here's the question:
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

Does this question mean that the numbers from 16 to y would be added and then see if it equals y? I have a hard time pinpointing what they're asking. When x/y it will yield the sum from 16 to y?

Please help.

Thanks,
KM

Hi KM,

Basically (x#y) is the remainder when x / y. You can think of it as two processes:
(1) What are all the numbers such that when divided by 16 results in a remainder of 1?
(2) What is the sum of all the numbers in (1)?

One way to think about it is to realize that if 16 / y has a remainder of 1, then 15 / y has a remainder of 0, or y is a factor of 15.

The factors of 15 are 1, 3, 5, and 15. If we use these values to evaluate 16 / y, we find that 3, 5, and 15 all work and result in a remainder of 1; 1 does not work.

So the possible values of y are 3, 5, and 15; the sum will be 23.

Hope that helps.