Question 1152784
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List out the various coins
{penny, nickel, dime, quarter, half dollar, silver dollar}
We are told that no coin is less than 3 cents, so we can cross off the penny
{<s>penny</s>, nickel, dime, quarter, half dollar, silver dollar}
leaving us with this set
{nickel, dime, quarter, half dollar, silver dollar}


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and we are also told that no coin is worth more than 30 cents, so we can cross off the half dollar and silver dollar coins
{nickel, dime, quarter, <s>half dollar</s>, <s>silver dollar</s>}
leaving us with
{nickel, dime, quarter}


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Therefore, the coins we're dealing with are: {nickel, dime, quarter}


n = number of nickels
d = number of dimes
q = number of quarters


n+d+q = 27 because we have 27 coins
Solve for q to get q = 27-n-d


5n = value, in cents, of all the nickels 
10d = value, in cents, of all the dimes
25q = value, in cents, of all the quarters
5n+10d+25q = total value of all the coins
total value = 365 cents = 3.65 dollars
5n+10d+25q = 365


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5n+10d+25q = 365
5n+10d+25(27-n-d) = 365 ... plug in q = 27-n-d
5n+10d+25(27)+25(-n)+25(-d) = 365 ... distribute
5n+10d+675-25n-25d = 365
-20n-15d+675 = 365
-20n-15d+675-675 = 365-675 ... subtract 675 from both sides
-20n-15d = -310 
-5(4n+3d) = -310 
4n+3d = 62 .... divide both sides by -5


Now solve for n
4n+3d = 62
4n = -3d+62
n = (-3d+62)/4
n = (-3d+2+60)/4
n = (-3d+2)/4+60/4
n = (-3d+2)/4+15
Through trial and error, you'll find that d = 2 makes the expression (-3d+2)/4 result in an integer
n = (-3d+2)/4+15
n = (-3*2+2)/4+15
n = (-6+2)/4+15
n = -4/4+15
n = -1+15
n = 14
So we have d = 2 dimes pair up with n = 14 nickels.


If d = 2 and n = 14, then,
q = 27-n-d
q = 27-2-14
q = 11


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Now increase d = 2 to d = 6. This is an increase of +4, which is from the denominator of 4 in (-3d+2)/4


We have,
n = (-3d+2)/4+15
n = (-3*6+2)/4+15
n = (-18+2)/4+15
n = -16/4+15
n = -4+15
n = 11


Now use n = 11 and d = 6 to get,
q = 27-n-d
q = 27-11-6
q = 10


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Two solutions are
(n,d,q) = (14,2,11)
and
(n,d,q) = (11,6,10)


For the solution (n,d,q) = (14,2,11), we can say
5n = 5*14 = 70 cents is from the nickels only
10d = 10*2 = 20 cents is from the dimes only
25q = 25*11 = 275 cents is from the quarters only
5n+10d+25q = 70+20+275 = 365 cents = $3.65 is the total combined value
This confirms the solution (n,d,q) = (14,2,11)


For the solution (n,d,q) = (11,6,10), we can say
5n = 5*11 = 55 cents is from the nickels only
10d = 10*6 = 60 cents is from the dimes only
25q = 25*10 = 250 cents is from the quarters only
5n+10d+25q = 55+60+250 = 365 cents = $3.65 is the total combined value
This confirms the solution (n,d,q) = (11,6,10)



There are 5 total ways to have nickels, dimes, and quarters combine to $3.65 such that we have 27 coins total
<table border = "1" cellpadding = "5">
<tr><td>Case</td><td>Nickels</td><td>Dimes</td><td>Quarters</td></tr>
<tr><td>A</td><td>2</td><td>18</td><td>7</td></tr>
<tr><td>B</td><td>5</td><td>14</td><td>8</td></tr>
<tr><td>C</td><td>8</td><td>10</td><td>9</td></tr>
<tr><td>D</td><td>11</td><td><font color=red>6</font></td><td>10</td></tr>
<tr><td>E</td><td>14</td><td><font color=red>2</font></td><td>11</td></tr>
</table>
Each row adds to 27.
The solutions we found earlier are in cases D and E of this table.


So Roger could have <font color=red>6 dimes</font> (case D) or he could have <font color=red>2 dimes</font> (case E).
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