We never see any half dollars anymore, so I will exclude them.
1) coins equal up to one dollar
25q+10d+5n+p=100
2) he has no more than 100 coins
q+d+n+p <= 100
3) he has 3 times as many quarters as dimes
q = 3d
4) he had the same number of nickels as quarters.
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:
1) coins equal up to one dollar,
since 100 pennies is one dollar
2) he has no more than 100 coins
100 pennies is no more than 100 coins
3) he has 3 times as many quarters as dimes
0 times 3 is 0
4) he had the same number of nickels as quarters.
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:
3) he has 3 times as many quarters as dimes
He has 3 quarters, and since
4) he had the same number of nickels as quarters.
He has 3 quarters.
So he has 3 quarters, 1 dime, and 3 nickels,
or the funny answer, 100 pennies.
Edwin