Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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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 12 - 7 = \boxed{5}. <math>\textbf{(C)}</math> | + | 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> |
==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 00:53, 5 February 2015
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 5 between 5 and 
inclusive can be obtained by a combination of coins. Thus, 
combinations can be made, so 
. But the answer is not 7, because we are asked for the number of 10-cent coins, which 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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