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
B/p + A/100 = (A + B/100)/p
B + Ap/100 = A + B/100
A(1 - p/100) = 99B/100
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.
(100-p)/p = n, where n is an integer.
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