Question 1032751
<pre>
We never see any half dollars anymore, so I will exclude them.
</pre>1) coins equal up to one dollar<pre>
25q+10d+5n+p=100</pre>
2) he has no more than 100 coins<pre>
q+d+n+p <= 100</pre>
3) he has 3 times as many quarters as dimes<pre>
q = 3d</pre>
4) he had the same number of nickels as quarters.<pre>
n = q

25q+10d+5n+p=100

Since q=3d and n=q, substitute 3d for q and n

25(3d)+10d+5(3d)+p=100

75d+10d+15d+p=100

100d+p=100

d could only be 0 or 1

If d=0 then he has 100 pennies.
Yes that's possible, because:</pre>
1) coins equal up to one dollar,<pre> 
since 100 pennies is one dollar</pre> 
2) he has no more than 100 coins<pre>
100 pennies is no more than 100 coins</pre>
3) he has 3 times as many quarters as dimes<pre>
0 times 3 is 0</pre>
4) he had the same number of nickels as quarters.<pre>
0 = 0

So that's one solution: 100 pennies, 
but it's a funny one!

If d = 1

100d+p = 100

100(1)+p = 100

100+p = 100

    p = 0

The other solution is 1 dime, no pennies and since:</pre>
3) he has 3 times as many quarters as dimes<pre>
He has 3 quarters, and since</pre>
4) he had the same number of nickels as quarters.<pre>
He has 3 quarters.

So he has 3 quarters, 1 dime, and 3 nickels,
or the funny answer, 100 pennies.

Edwin</pre>