Question 561946
With a systematic analysis I find 4 solutions and can be pretty sure that I have all the solutions.
Let's define variables
{{{q}}} = number of quarters
{{{d}}} = number of dimes
{{{n}}} = number of nickels
{{{p}}} = number of pennies
There are 21 coins, so {{{q+d+n+p=21}}}
The value adds to $1.00, so {{{25q+10d+5n+p=100}}} --> {{{5q+2d+n+p/5=20}}}
So {{{p}}} must be a multiple of 5, meaning that it must be 0, 5, 10, 15, or 20.
It cannot be 0, because the 21 coins would add up to at least $1.05 (21 nickels).
It cannot be 20 pennies, because the remaining coin would have to be worth $0.80.
FIVE PENNIES
Wirh 5 pennies the remaining 16 coins add up to 95 cents, so we have
{{{25q+10d+5n=95}}} --> {{{5q+2d+n=19}}} and
{{{q+d+n=16}}}
Combining both equations we get {{{4q+d=3}}}.
The only integer solution is {{{q=0}}}, {{{d=3}}}
Substituting in {{{q+d+n=16}}}
we get {{{3+n=16}}} --> {{{n=13}}}
First solution: no quarters, 3 dimes, 13 nickels, 5 pennies
TEN PENNIES
With 10 pennies the remaining 11 coins add up to 90 cents, so we have
{{{25q+10d+5n=90}}} --> {{{5q+2d+n=18}}} and
{{{q+d+n=11}}}
Combining both equations we get {{{4q+d=7}}}.
The integer solutions are {{{q=1}}}, with {{{d=3}}}
and {{{q=0}}}, with {{{d=7}}}
For {{{q=1}}}, with {{{d=3}}},
{{{q+d+n=11}}} means {{{4+n=11}}} --> {{{n=7}}}
For {{{q=0}}}, with {{{d=7}}},
{{{q+d+n=11}}} means {{{7+n=11}}} --> {{{n=4}}}
Two more solutions:
1 quarter, 3 dimes, 7 nickels, 10 pennies
no quarters, 7 dimes, 4 nickels, 10 pennies
FIFTEEN PENNIES
With 15 pennies the remaining 6 coins add up to 85 cents, so we have
{{{25q+10d+5n=85}}} --> {{{5q+2d+n=17}}} and
{{{q+d+n=6}}}
Combining both equations we get {{{4q+d=11}}}.
The integer solutions are {{{q=2}}}, with {{{d=3}}},
{{{q=1}}}, with {{{d=7}}},
and {{{q=0}}}, with {{{d=11}}}
However, only the first solution works with {{{q+d+n=6}}}
and gives us {{{5+n=6}}} --> {{{n=1}}}
One more solutions:
2 quarters, 3 dimes, 1 nickel, 15 pennies