Question 1105371: Helen has some $1, $2 and $5 coins. The total value is $80. All the $1 coins may be traded in for $10 coins, resulting in 36 fewer coins. All the $5 coins may be traded in for $10 coins, and all the $2 coins may be traded in for $5 coins.
What is the largest possible numbers of coins Helen has?
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
Let x, y, and z be the numbers of $1, $2, and $5 coins, respectively.
We then know 
(1) If exchanging all the $1 coins for $10 coins results in 36 fewer coins, then the number of $1 coins is 40.
So now we know  , so  .
(2) Since y and z are positive integers, y must be a multiple of 5 and z must be even.
(3) Since we are looking for the largest number of coins she can have, we want y to be as large as possible and z to be as small as possible.
(4) The smallest even positive integer is 2, so z should be 2; that makes y = 15.
The largest possible number of coins is 57, with 40 $1 coins, 15 $2 coins, and 2 $5 coins.
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