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?

Initial Amount in dollars = A + B/100

Final Amount in dollars = B/p + A/100

Final Amount = Initial Amount/p

Therefore,

B/p + A/100 = (A + B/100)/p

Therefore,

B + Ap/100 = A + B/100

Therefore,

A(1 - p/100) = 99B/100

Therefore,

A = 99B/(100 - p)

But A and B are integers.

Therefore, either 99/(100-p) is an integer OR B/(100-p) is an integer.

Let us first consider the case that 99/(100-p) is an integer. This happens when (100-p) is either 1,11,3,9,33 or 99. That is, factors of 99. This in turn yields p = 99 or 89 or 97 or 81 or 67. (note 1 is excluded since p is between 2 and 99)

Then using the formula:

A = 99B/(1-p)

we arrive at A = B (for p=99), 9B (for p =89), 33B (p=97), 11B (p=81) or 3B (p=67) or

Using the condition that B/p is an integer, and assuming B <99, then B=p for each of the above, since 2*p exceeds 99. But if we set B = p for the above p values and evaluate A = 99B/(1-p) we exceed 99. Therefore these p values are not valid.

To find the remaining possible values for p, we consider the case where B/(100-p) is an integer.

Now since A = 99 * B/(100-p), if B/(100-p) > 1 then A will exceed 99. Therefore B/(100-p) = 1, so B = (100-p).

Also we require that B/p is an integer. Therefore we require that (100-p)/p be an integer for this case.

That is,

(100-p)/p = n, where n is an integer.

Therefore,

100 = p(n+1)

Therefore, for this case p must be factors of 100.

The factors of 100 include 50,2,25,4,20,5,10.

Therefore answer is p =50 or 2 or 25 or 4 or 20 or 5 or 10.

*Edited on ***August 13, 2012, 10:25 am**

*Edited on ***August 13, 2012, 10:32 am**