Question 285547
The average value would be
(total value)/(number of coins)
Let {{{q}}} = number of quarters
Let {{{d}}} = number of dimes
Let {{{n}}} = number of nickels
Let {{{p}}} = number of pennies
given:
(1) {{{(25q + 10d + 5n + p)/(q + d + n + p) = 17}}} (in cents)
(1) {{{25q + 10d + 5n + p = 17*(q + d + n + p)}}}
and
{{{(25q + 10d + 5n + 1*(p - 1))/(q + d + n + p - 1) = 18}}}
{{{25q + 10d + 5n + p - 1 = 18*(q + d + n + p - 1)}}}
(2) {{{25q + 10d + 5n + p = 18*(q + d + n + p - 1) + 1}}}
Comparing (1) and (2) I can say
{{{17*(q + d + n + p) = 18*(q + d + n + p - 1) + 1}}}
{{{17*(q + d + n + p) = 18*(q + d + n + p) - 17}}}
Subtract {{{17*(q + d + n + p)}}} from both sides
{{{0 = q + d + n + p - 17}}}
{{{q + d + n + p = 17}}}
Going back to (1),
(1) {{{(25q + 10d + 5n + p)/(q + d + n + p) = 17}}}
(1) {{{(25q + 10d + 5n + p)/17 = 17}}}
{{{25q + 10d + 5n + p = 289}}} (in cents)
There must be {{{17}}} coins, making {{{289}}} cents 
I know there must be at least {{{4}}} pennies
and there could be {{{9}}} or {{{14}}} pennies
There can't be {{{12}}} quarters, since that adds to {{{300}}} cents
There can't be {{{11}}} quarters either, since {{{1}}} dime and
{{{1}}} nickel make {{{275 + 10 + 5 = 290}}} -too much
Suppose there are {{{10}}} quarters, {{{1}}} dime, {{{1}}} nickel,
and {{{4}}} pennies
That makes {{{250 + 10 + 5 + 4 = 269}}}, but {{{10 + 1 + 1 + 4 = 16}}} coins
and I can't make up {{{20}}} cents with {{{1}}} more coin
I'm guessing there are {{{9}}} quarters -you can figure out the rest