This is a generalization of Rupees and Paise.

Stan entered a departmental store with A dollars and B cents. When he exited the store, he had B/p dollars and A cents, where B/p is an integer. It was observed that when Stan came out, he had precisely 1/p times the money he had when he came in.

Given that each of A, B and p is a positive integer, with 2 ≤ p ≤ 99, determine the values of p for which this is possible. What values of p generate more than one solution?

Suppose A=99, B=100-p, and p|100. Let n=100/p. Then, B/p=(100-p)/p=n-1. Then, (100A+B)/p=(100(99)+100-p)/p=(9900+100-p)/p=(10000-p)/p=100n-1=100n-100+99=100(n-1)+99=100(B/p)+A. Therefore, we have the following solutions.

Posted by Math Man on 2020-05-16 11:55:29