Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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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 | 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 | ||
<math>\boxed{\textbf{(C) } 5}</math> | <math>\boxed{\textbf{(C) } 5}</math> | ||
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==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}} |
Revision as of 11:50, 11 April 2015
Contents
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?
Solution
Let Claudia have 5-cent coins and 10-cent coins. It is easily observed that any multiple of between and inclusive can be obtained by a combination of coins. Thus, combinations can be made, so . But the answer is not because we are asked for the number of 10-cent coins, which is
Solution #2
Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of To have exactly different multiples of we will need to make up to cents. If all twelve coins were 5-cent coins, we will have cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain cents, and as we need to gain cents, the answer is
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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