Im an idiot. Just answered by own question as 1 goes into 16 evenly with no remainder. Sorry!

Just in case anyone else was wondering about this, we have:

(16/y) has remainder 1, or

16 = y*(something) + 1, or

15 = y*(something)

So y is a factor of 15 other than 1 itself. (Remember that any positive integer divided by 1 has remainder 0.)

Hence y = 3, 5, or 15, and the sum of solutions = 3 + 5 + 15 = 23.

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


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
==> (x#y)= the remainder after dividing x by y. (16#y)=1 means that the remainder is 1 when you divide 16 by y. Since 16=2^4, dividing it by an odd number would mean the remainder is 1 (except 1). From 16=3*5+1=5*3+1=15*1+1, y=3,5,15 and the sum would be 3+5+15=23. Therefore the answer is D



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