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Dollars and Cents (Posted on 2012-08-13) Difficulty: 3 of 5
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?

See The Solution Submitted by K Sengupta    
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PARTIAL answer Comment 5 of 5 |

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                                                   


  Posted by Ady TZIDON on 2012-08-14 07:47:43
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