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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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Solution re: Solution | Comment 2 of 5 |
(In reply to Solution by Chris, PhD)

In my previous post, I made the assumption that B and A are smaller than or equal to 99.  I made this assumption because I thought that extra cents would be rounded to dollars in the question.

If this isn't a correct assumption to make then the p values I mentioned for when 99/(1-p) is an integer also are valid.  These solutions also give multiple answers for A and B:

We arrive at A = B (for p=99), 9B (for p =91), 33B (p=97), 11B (p=89) or 3B (p=67).

Where all values of B are possible for which B/p is an integer.

I will show that these p values are also valid:
p =50 or 2 or 25 or 4 or 20 or 5 or 10.

Consider B/(100-p) is an integer and B/p is an integer.
I.e. B/(100-p) = n and B/p = k, where n and k are integers.

Therefore B = n(100-p) and B = pk.

Therefore,

n(100-p) = pk

Therefore,

p(k+n) = 100n.

Therefore p includes all factors of 100 as per the previous solution.

Have to ponder on this.  But taking the assumption that B and A are not restricted by 99, I have found that:
p = 99,91,97,89,67 (these provide multiple solutions for A and B)
and
p = 50,2,25,4,20,5,10.


Edited on August 13, 2012, 10:36 pm
  Posted by Chris, PhD on 2012-08-13 10:48:21

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