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Difference between revisions of "2015 AMC 10A Problems/Problem 13"

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<math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math>
 
<math> \textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7 </math>
  
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==Solution 1==
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Let Claudia have <math>x</math> 5-cent coins and <math>\left( 12 - x \right)</math> 10-cent coins. It is easily observed that any multiple of <math>5</math> between <math>5</math> and <math>5x + 10(12 - x) = 120 - 5x</math> inclusive can be obtained by a combination of coins. Thus, <math>24 - x = 17</math> combinations can be made, so <math>x = 7</math>. But the answer is not <math>7,</math> because we are asked for the number of 10-cent coins, which is <math>12 - 7 = \boxed{\textbf{(C) } 5}</math> kurt
  
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==Solution==
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==Solution 2==
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Let Claudia have <math>x</math> 5-cent coins and <math>12-x</math> 10-cent coins. It is easily observed that any multiple of 5 between 5 and <math>5x + 10(12 - x) = 120 - 5x</math> inclusive can be obtained by a combination of coins. Thus, <math>24 - x = 17</math> combinations can be made, so <math>x = 7</math>. But the answer is not 7, because we are asked for the number of 10-cent coins, which is <math>12 - 7 = \boxed{5}</math>. <math>\textbf{(C)}</math>
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Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of <math>5.</math> To have exactly <math>17</math> different multiples of <math>5,</math> we will need to make up to <math>85</math> cents. If all twelve coins were 5-cent coins, we will have <math>60</math> cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain <math>5</math> cents, and as we need to gain <math>25</math> cents, the answer is
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<math>\boxed{\textbf{(C) } 5}</math>
  
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==Solution 3 (Quick Insight)==
  
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==Alternate Solution==
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Notice that for every <math>d</math> dimes, any multiple of <math>5</math> less than or equal to <math>10d + 5(12-d)</math> is a valid arrangement. Since there are <math>17</math> in our case, we have <math>10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5</math>. Therefore, the answer is <math>\boxed{\textbf{(C) } 5}</math>.
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Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of 5. To have exactly 17 different multiples of 5, we will need to make up to 85 cents. If all twelve coins were 5-cent coins, we will have 60 cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain 5 cents, and as we need to gain 25 cents, the answer is  
 
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<math>\boxed{5}</math>.  <math>\textbf{(C)}</math>
 
  
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~MrThinker
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== Solution 4 ==
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Dividing by 5cents to reduce clutter:
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<math>D </math> double coins and <math>S </math> single coins can reach any value between <math>1</math> and <math>2D + S</math>. Set <math>2D + S = 17 </math> and <math>D + S = 12</math>.
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Subtract to get <math>D = \boxed{\textbf{(C) } 5}</math>.
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~oinava
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==Video Solution==
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https://youtu.be/F2iyhLzmCB8
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2015|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 08:28, 31 October 2024

Problem 13

Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution 1

Let Claudia have $x$ 5-cent coins and $\left( 12 - x \right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$. But the answer is not $7,$ because we are asked for the number of 10-cent coins, which is $12 - 7 = \boxed{\textbf{(C) } 5}$ kurt

Solution 2

Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of $5.$ To have exactly $17$ different multiples of $5,$ we will need to make up to $85$ cents. If all twelve coins were 5-cent coins, we will have $60$ cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain $5$ cents, and as we need to gain $25$ cents, the answer is $\boxed{\textbf{(C) } 5}$

Solution 3 (Quick Insight)

Notice that for every $d$ dimes, any multiple of $5$ less than or equal to $10d + 5(12-d)$ is a valid arrangement. Since there are $17$ in our case, we have $10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5$. Therefore, the answer is $\boxed{\textbf{(C) } 5}$.

~MrThinker

Solution 4

Dividing by 5cents to reduce clutter:


$D$ double coins and $S$ single coins can reach any value between $1$ and $2D + S$. Set $2D + S = 17$ and $D + S = 12$.

Subtract to get $D = \boxed{\textbf{(C) } 5}$.

~oinava

Video Solution

https://youtu.be/F2iyhLzmCB8

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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