Difference between revisions of "2015 AMC 10A Problems/Problem 13"
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~MrThinker | ~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== | ==Video Solution== | ||
https://youtu.be/F2iyhLzmCB8 | https://youtu.be/F2iyhLzmCB8 |
Revision as of 14:36, 11 July 2024
Contents
[hide]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 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
Solution 4
Dividing by 5cents to reduce clutter:
double coins and single coins can reach any value between and . Set and .
Subtract to get .
~oinava
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 |
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