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
-so my post contributes nothing new
PARTIAL answer
