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?

Given: (100*A+B)/ (100*B+A/p) =p

A*(100-p)=99*B

B =A*(100-p) /99 <o:p></o:p>

Case 1 A=99

When A=99 B =100-p, valid only for values of p that divide 100-p

Those are: 1(Trivial A=B), 5,10,20,25,50,

100 (N.A,leading to A=B/0 }

Answers : (p,A,B)= (2,99,98) ; (4,99,96) (5,99,95); (10,99,90); (20,99,80);

(25,99,75);(50 ,99, 50);-<o:p></o:p>

Leaving the store with 49.99; 24.99 19.99; 9.99; 9.99; 4.99; 3.99; 2.99; respectively

<o:p> </o:p>

a general approach: 99 divides

A*(100-p) and there exist integer values for

the appropriate A other that 99) & B (!):

excel table provides the following values for (p, (100-p) /99) :

rem : **Stopped copying due to :**

**a) format problems **

**b) read Charlie's post **

**-s****o my post contributes ****nothing new**