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How many different combinations of quarters, nickels, and dimes can be used to make $0.55?

Guest Apr 21, 2015

Best Answer 

 #3
avatar+20038 
+5

$$\small{\text{Quater $ = 25 \qquad $Dime $ = 10 \qquad $ Nickel $= 5
$}} \\
\small{\text{In \$0.55 max. 2 Quaters, max. 5 Dimes and max. 11 Nickels }} \\\\
\left(\sum\limits_{i=0}^{2} x^{(\textcolor[rgb]{0,0,1}{25}*i}) \right) \times \left (\sum\limits_{i=0}^{5} x^{(\textcolor[rgb]{0,0,1}{10}*i)} \right) \times \left(\sum\limits_{i=0}^{11} x^{(\textcolor[rgb]{0,0,1}{5}*i)} \right) \\\\
\small{\text{$=(1+x^{25}+x^{50})$}} \times
\small{\text{$(1+x^{10}+x^{20}+x^{30}+x^{40}+x^{50})$}} \\ \times
\small{\text{$(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}+x^{40}+x^{45}+x^{50}+x^{55})$}}\\
\small{\text{$
=(x^{155})+(x^{150})+2*x^{145}+2*x^{140}+3*x^{135}+4*x^{130} $}} \\
\small{\text{$+5*x^{125}+6*x^{120}+7*x^{115}+8*x^{110}+10*x^{105}+11*x^{100} $}} \\
\small{\text{$+11*x^{95}+12*x^{90}+12*x^{85}+13*x^{80}+13*x^{75}+12*x^{70} $}} \\
\small{\text{$+12*x^{65}+11*x^{60}+\textcolor[rgb]{1,0,0}{11*x^{55}}+10*x^{50}+8*x^{45}+7*x^{40} $}} \\
\small{\text{$+6*x^{35}+5*x^{30}+4*x^{25}+3*x^{20}+2*x^{15}+2*x^{10}+(x^5)+1 $}} \\$$

 

The coefficient from  $$\small{\text{$\textcolor[rgb]{1,0,0}{x^{55}}$}}$$ is $$\small{\text{$\textcolor[rgb]{1,0,0}{ 11 }$}}$$. So there are 11 possibilities.

heureka  Apr 21, 2015
 #1
avatar+90180 
+5

Let's see.......

2 quarters  1 nickel

1 quarter  3 dimes

1 quarter 2 dimes  2 nickels

1 quarter  1 dime 4 nickels

1 quarter 6 nickels

5 dimes 1 nickel

4 dimes 3 nickels

3 dimes 5 nickels

2 dimes 7 nickels

1 dime 9 nickels

11 nickels

 

I think that's it......did I leave anything  out ???

 

  

CPhill  Apr 21, 2015
 #2
avatar+20038 
+5

 

How many different combinations of quarters, nickels, and dimes can be used to make $0.55 ?

 

 $$\small{\text{
$
\begin{array}{rccrlclcl}
\hline
\\
1. & \$0.55 &=& & && 3\; Dimes &+& 1\; Quater \\
2. & \$0.55 &=& 1& Nickel && &+& 2\; Quaters \\
3. & \$0.55 &=& 1& Nickel &+& 5\; Dimes \\
4. & \$0.55 &=& 2& Nickels &+& 2\; Dimes &+& 1\; Quater \\
5. & \$0.55 &=& 3& Nickels &+& 4\; Dimes \\
6. & \$0.55 &=& 4& Nickels &+& 1\; Dime &+& 1\; Quater\\
7. & \$0.55 &=& 5& Nickels &+& 3\; Dimes \\
8. & \$0.55 &=& 6& Nickels && &+& 1\; Quater \\
9. & \$0.55 &=& 7& Nickels &+& 2\; Dimes \\
10. & \$0.55 &=& 9& Nickels &+& 1\; Dime \\
11. & \$0.55 &=& 11& Nickels \\
\\
\hline
\end{array}
$}}$$

heureka  Apr 21, 2015
 #3
avatar+20038 
+5
Best Answer

$$\small{\text{Quater $ = 25 \qquad $Dime $ = 10 \qquad $ Nickel $= 5
$}} \\
\small{\text{In \$0.55 max. 2 Quaters, max. 5 Dimes and max. 11 Nickels }} \\\\
\left(\sum\limits_{i=0}^{2} x^{(\textcolor[rgb]{0,0,1}{25}*i}) \right) \times \left (\sum\limits_{i=0}^{5} x^{(\textcolor[rgb]{0,0,1}{10}*i)} \right) \times \left(\sum\limits_{i=0}^{11} x^{(\textcolor[rgb]{0,0,1}{5}*i)} \right) \\\\
\small{\text{$=(1+x^{25}+x^{50})$}} \times
\small{\text{$(1+x^{10}+x^{20}+x^{30}+x^{40}+x^{50})$}} \\ \times
\small{\text{$(1+x^{5}+x^{10}+x^{15}+x^{20}+x^{25}+x^{30}+x^{35}+x^{40}+x^{45}+x^{50}+x^{55})$}}\\
\small{\text{$
=(x^{155})+(x^{150})+2*x^{145}+2*x^{140}+3*x^{135}+4*x^{130} $}} \\
\small{\text{$+5*x^{125}+6*x^{120}+7*x^{115}+8*x^{110}+10*x^{105}+11*x^{100} $}} \\
\small{\text{$+11*x^{95}+12*x^{90}+12*x^{85}+13*x^{80}+13*x^{75}+12*x^{70} $}} \\
\small{\text{$+12*x^{65}+11*x^{60}+\textcolor[rgb]{1,0,0}{11*x^{55}}+10*x^{50}+8*x^{45}+7*x^{40} $}} \\
\small{\text{$+6*x^{35}+5*x^{30}+4*x^{25}+3*x^{20}+2*x^{15}+2*x^{10}+(x^5)+1 $}} \\$$

 

The coefficient from  $$\small{\text{$\textcolor[rgb]{1,0,0}{x^{55}}$}}$$ is $$\small{\text{$\textcolor[rgb]{1,0,0}{ 11 }$}}$$. So there are 11 possibilities.

heureka  Apr 21, 2015

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