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Pennies, Dimes, and Quarters (Posted on 2009-02-17) Difficulty: 2 of 5
In my left pocket I have a mix of pennies (1 cent) and quarters (25 cents). In my right pocket I have a bunch of dimes (10 cents). The number of coins in each pocket is the same, so is the cash value. What is the smallest (nonzero) number of coins I can have?

What if I had coins of x cents and z cents in my left pocket and coins of y cents in my right pocket - is there some quantitiy of coins I can have so that each pocket has the same number of coins and same cash value? (The value of the coins are positive integers x > y > z.)

See The Solution Submitted by Brian Smith    
Rating: 3.5000 (2 votes)

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Solution the rest | Comment 5 of 11 |

Assume I had   A coins of x cents and   B    of  z cents in my left pocket and      A+B coins of y cents in my right pocket - then

Ax+Bz=(A+B)Y

A(x-y)=B(y-z)

Clearly  A=y-z    and  B=  x-y ARE VALID POSITIVE SOLUTIONS

 

 


  Posted by Ady TZIDON on 2009-02-17 13:02:49
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