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 Pennies, Dimes, and Quarters (Posted on 2009-02-17)
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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 Solution | Comment 8 of 10 |
Alright, so the first part was pretty simple.   I first made the assumption that since dimes always add up to 10s and quarters are 5s for the ones digit, the number of pennies must be divisible by 5.   Starting at 5 pennies and 1 quarter, that's 30 cents or 3 dimes, which is not the answer.   Next test was 5 pennies and 3 quarters which was 80 cents, or 8 dimes.   DING!   16 coins.   The lowest number of coins that includes 10 pennies would be 10 and 2 quarters which is 12 coins which would be a minimum of 24 coins and that is larger than 16 so we've found our answer.

The second part is once again easy and in fact easier with the information you have given.   Since you put no restriction on the size of the coinage, just that they have to be positive and x>y>z, the lowest possible values are either 2>1>0 or 3>2>1.   In both cases you have 1 x coin, 1 z coin and 2 y coins.   As this site(http://mathworld.wolfram.com/PositiveInteger.html) clearly states that positive integers do not include the number zero(I couldn't remember so I looked it up), that means the answer is x=3, y=2, z=1. 4 cents in each pocket, 4 coins total.   That of course is the lowest possible value.   Any value where x+z=2y would work just as well and getting more complicated, any value where Ax+Bz=Cy(where A, B, and C are numbers of coins, and x, y, and z are coin values) would work too.

 Posted by Brandon on 2009-02-17 15:00:13

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