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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: 2.0000 (1 votes)

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Some Thoughts On the generalization | Comment 7 of 10 |
This is really a beautiful symmetrical result.  The other posts show the algebra, but allow me to show how simple it becomes in practice:

The number of each type of coins is just the difference of the values of the other two (provided you then divide out the gcd of these differences.)

For values 1, 25, 10 The differences:
25-10=15 pennies
10-1=9 quarters
25-1=24 dimes
but since 15, 9 and 24 are all divisible by 3 we get

5 pennies, 3 quarters, 24 dimes.



  Posted by Jer on 2009-02-17 14:39:33
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