Question 1033082
Four unknown variables and two equations.  The easiest restriction is on how many quarters, being some natural number from 1 to 4.


p, n, d, q the count of each coin.
{{{0.01p+0.05n+0.1d+0.25q=1.05}}}
{{{p+5n+10d+25q=105}}} from account of the money value


{{{p+n+d+q=14}}} the account of the coins



Solve each for p and equate.
{{{system(p=105-5n-10d-25q,p=14-n-d-q)}}}

{{{105-5n-10d-25q=14-n-d-q}}}
{{{-5n-10d-25q+n+d+q=14-105}}}
{{{-4n-9d-24q=-91}}}
{{{highlight_green(4n+9d+24q=91)}}}


CASES FOR THE QUARTERS
q=4
{{{4n+9d=91-24*4}}}
{{{4n+9d=-5}}}
-
q=3
{{{4n+9d=91-24*3}}}
{{{4n+9d=19}}}
-
q=2
{{{4n+9d=43}}}
-
q=1
{{{4n+9d=66}}}
-
Which of these cases will work to give the right whole number values for n and d?



Look at a graph for case q=1.  Treat vertical axis as d, horizontal as n,
looks bad!
{{{graph(300,300,-1,15,-1,15,-(4/9)x+66/9)}}}


The graph n and d for case q=2,
guess is that still not good.
{{{graph(300,300,-1,12,-1,12,-4x/9+43/9)}}}


See for case q=3,
{{{graph(300,300,-1,10,-1,10,-4x/9+19/9)}}}
Maybe....


A different way could be play with the simplest combination of coins to give $1.05 and split some coins to get a count of 14.
{{{system(q=4,n=1,c=5)}}}, c for count of coins but we must have c=14.  Splitting the nickel might not help much, at least not yet.


Split one quarter.
{{{system(q=3,d=2,n=2,c=7)}}}----still not enough coins c.


Split one more quarter.
{{{system(q=2,d=4,n=3,c=9)}}}----we want to be able to get an even number of coins.


Maybe split one nickel from that.
{{{system(q=2,d=4,n=2,p=5,c=13)}}}-----we want ONE more coin, so maybe split one dime.


{{{system(Try then q=2,d=3,n=4,p=5,c=14)}}}-----THIS must be the combination.
2 quarters
3 dimes
4 nickels
5 pennies
--That is 14 coins.