Question 1105371
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Let x, y, and z be the numbers of $1, $2, and $5 coins, respectively.<br>
We then know {{{x+2y+5z = 80}}}<br>
(1) If exchanging all the $1 coins for $10 coins results in 36 fewer coins, then the number of $1 coins is 40.<br>
So now we know {{{40+2y+5z = 80}}}, so {{{2y+5z = 40}}}.<br>
(2) Since y and z are positive integers, y must be a multiple of 5 and z must be even.<br>
(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.<br>
(4) The smallest even positive integer is 2, so z should be 2; that makes y = 15.<br>
The largest possible number of coins is 57, with 40 $1 coins, 15 $2 coins, and 2 $5 coins.