Question 164168
<pre><font size = 4 color = "indigo"><b>

I'll just work the problem for you. You'll have to
write it up using complete sentences and proper grammar,
and filling in your own explanations.

Let p = the number of pennies
Let n = the number of nickels
Let d = the number of dimes

We have the equation:

{{{.01p + .05n + .10d = 1.65}}}

or multiplying through by 100,

{{{p+5n+10d=165}}}

Since half the coins are nickels, we have

{{{n = (p+n+d)/2}}}

{{{2n = p+n+d}}}

{{{n=p+d}}} 

We have the system:

{{{system(p + 5n + 10d = 165,n=p+d)}}}

Substituting {{{p+d}}} for {{{n}}} in the first equation

{{{p + 5(p+d) + 10d = 165}}}

{{{p + 5p+5d + 10d = 165}}}

{{{6p + 15d = 165}}}

Divide through by 3:

{{{2p + 5d = 55}}}

{{{2p = 55-5d}}}

{{{2p = 5(11-d)}}}

Since the left side is even,
the right side must be even also,
and positive. Therefore {{{11-d}}} 
must be even and positive, too, 
Therefore d must be odd and no more 
than 11.

Therefore there are 6 possibilities:

d = 1, 3, 5, 7, 9, or 11

If {{{d=1}}}, then

{{{2p = 5(11-d)}}} becomes
{{{2p = 5(11-1)}}}
{{{2p = 5(10)}}}
{{{2p = 50}}}
{{{p = 25}}}

Substituting {{{d=1}}} and {{{p=25}}}
into

{{{p + 5n + 10d = 165}}}
{{{25 + 5n + 10(1) = 165}}}
{{{25 + 5n + 10 = 165}}}
{{{35 + 5n = 165}}}
{{{5n = 130}}}
{{{n = 26}}}

So one possibility is 

{{{p=25}}}, {{{n=26}}}, {{{d=1}}}

-------

If {{{d=3}}}, then

{{{2p = 5(11-d)}}} becomes
{{{2p = 5(11-3)}}}
{{{2p = 5(8)}}}
{{{2p = 40}}}
{{{p = 20}}}

Substituting {{{d=3}}} and {{{p=20}}}
into

{{{p + 5n + 10d = 165}}}
{{{20 + 5n + 10(3) = 165}}}
{{{20 + 5n + 30 = 165}}}
{{{50 + 5n = 165}}}
{{{5n = 115}}}
{{{n = 23}}}

So another possibility is 

{{{p=20}}}, {{{n=23}}}, {{{d=3}}}

-----------------

If {{{d=5}}}, then

{{{2p = 5(11-d)}}} becomes
{{{2p = 5(11-5)}}}
{{{2p = 5(6)}}}
{{{2p = 30}}}
{{{p = 15}}}

Substituting {{{d=5}}} and {{{p=15}}}
into

{{{p + 5n + 10d = 165}}}
{{{15 + 5n + 10(5) = 165}}}
{{{15 + 5n + 50 = 165}}}
{{{65 + 5n = 165}}}
{{{5n = 100}}}
{{{n = 20}}}

So another possibility is 

{{{p=15}}}, {{{n=20}}}, {{{d=5}}}

-------------

If {{{d=7}}}, then

{{{2p = 5(11-d)}}} becomes
{{{2p = 5(11-7)}}}
{{{2p = 5(4)}}}
{{{2p = 20}}}
{{{p = 10}}}

Substituting {{{d=7}}} and {{{p=10}}}
into

{{{p + 5n + 10d = 165}}}
{{{10 + 5n + 10(7) = 165}}}
{{{10 + 5n + 70 = 165}}}
{{{80 + 5n = 165}}}
{{{5n = 85}}}
{{{n = 17}}}

So another possibility is 

{{{p=10}}}, {{{n=17}}}, {{{d=7}}}

-------------

If {{{d=9}}}, then

{{{2p = 5(11-d)}}} becomes
{{{2p = 5(11-9)}}}
{{{2p = 5(2)}}}
{{{2p = 10}}}
{{{p = 5}}}

Substituting {{{d=9}}} and {{{p=5}}}
into

{{{p + 5n + 10d = 165}}}
{{{5 + 5n + 10(9) = 165}}}
{{{5 + 5n + 90 = 165}}}
{{{95 + 5n = 165}}}
{{{5n = 70}}}
{{{n = 14}}}

So another possibility is 

{{{p=5}}}, {{{n=14}}}, {{{d=9}}}

-------------

If {{{d=11}}}, then

{{{2p = 5(11-d)}}} becomes
{{{2p = 5(11-11)}}}
{{{2p = 5(0)}}}
{{{2p = 0}}}
{{{p = 0}}}

Substituting {{{d=11}}} and {{{p=0}}}
into

{{{p + 5n + 10d = 165}}}
{{{0 + 5n + 10(11) = 165}}}
{{{0 + 5n + 110 = 165}}}
{{{5n = 55}}}
{{{n = 11}}}

So another possibility is 

{{{p=0}}}, {{{n=11}}}, {{{d=11}}}

-------------

So all the 6 possibilities are 

0 pennies, 11 nickels, 11 dimes, 22 coins, half of which are 11 nickels.
5 pennies, 14 nickels, 9 dimes, 28 coins, half of which are 14 nickels.
10 pennies, 17 nickels, 7 dimes, 34 coins, half of which are 17 nickels.
15 pennies, 20 nickels, 5 dimes, 40 coins, half of which are 20 nickels.
20 pennies, 23 nickels, 3 dimes, 46 coins, half of which are 23 nickels.
25 pennies, 26 nickels, 1 dimes, 52 coins, half of which are 26 nickels.

Edwin</pre>