Question 668799

<pre>
I'll assume you don't mean to use pennies or half dollars

then let q = the number of quarters
     let d = the number of dimes
    then 3d = the number of nickels.

Then there is a "coin equation" and a "money equation":


The "coin" equation is

 {{{(matrix(4,1,

     The,number, of, quarters))}}} + {{{(matrix(4,1,

     The,number, of, dimes))}}} + {{{(matrix(4,1,

     The,number, of, nickels))}}} = 85


The "money" equation is:

{{{(matrix(6,1,

     "$0.25",times,the,number, of, quarters))}}} + {{{(matrix(6,1,

     "$0.10",times,the,number, of, dimes))}}} + {{{(matrix(6,1,

     "$0.05",times,the,number, of, nickelss))}}} = $6.25


{{{system(q+d+3d=85,0.25q+0.10d+0.05(3d)=6.25)}}}
Simplify:
{{{system(q+4d=85,0.25q+0.10d+0.15d=6.25)}}}
Simplify some more:
{{{system(q+4d=85,0.25q+0.25d=6.25)}}}
Multiply the second one through by 100 to clear of decimals:
{{{system(q+4d=85,25q+25d=625)}}}
Divide the second equation through by 25
{{{system(q+4d=85,q+d=25)}}}
  
Solve that and get q=5 and d=20

So there are 5 quarters, 20 dimes, and since there are 3d nickels,
there are 3(20) or 60 nickels.

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If you allow pennies there are two more solutions:

12 quarters, 12 dimes, 36 nickels and 25 pennies

19 quarters, 4 dimes, 12 nickels and 50 pennies.

But there are no solutions using half dollars.

Edwin</pre>