Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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<math>\boxed{\textbf{(C) } 5}</math> | <math>\boxed{\textbf{(C) } 5}</math> | ||
| + | ==Solution 3 (Quick Insight)== | ||
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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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| + | ~MrThinker | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/F2iyhLzmCB8 | https://youtu.be/F2iyhLzmCB8 | ||
Revision as of 18:31, 10 August 2023
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 his coins. How many 10-cent coins does Claudia have?
Solution 1
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
Solution 3 (Quick Insight)
Notice that for every 
dimes, any multiple of less than or equal to 
is a valid arrangement. Since there are in our case, we have 
. Therefore, the answer is .
~MrThinker
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
