2015 AMC 10A Problems/Problem 13
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 kurt
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 | |
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All AMC 10 Problems and Solutions |
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