2000 AMC 8 Problems/Problem 20
Contents
Problem
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $, with at least one coin of each type. How many dimes must you have?
Solution 1
Since you have one coin of each type, cents are already determined, leaving you with a total of cents remaining for coins.
You must have more penny. If you had more than penny, you must have at least pennies to leave a multiple of for the nickels, dimes, and quarters. But you only have more coins to assign.
Now you have cents remaining for coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves cents in nickels or quarters, which is impossible. If you have two dimes, that leaves cents for nickels or quarters, which is again impossible. If you have three dimes, that leaves cents for nickel or quarter, which is still impossible. And all four remaining coins being dimes will not be enough.
Therefore, you must have no more dimes to assign, and the cents in coins must be divided between the quarters and nickels. We quickly see that nickels and quarters work. Thus, the total count is quarters, nickels, penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total coins, and a total of cents.
There is only dime in that combo, so the answer is .
Solution 2 (Faster)
We see that there must be 102 cents, so therefore there's at least 2 pennies. That leaves 7 coins. We assume that there are 3 quarters, leaving 25 cents with 4 coins left. If all 4 are nickels, that would only be 20 cents, missing 5. Therefore, one nickel must be changed into 1 dime, so the answer is
Solution by ILoveMath31415926535
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=1409
~ pi_is_3.14
Video Solution 2
https://youtu.be/Vm-proRV5wI. Soo, DRMS, NM
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AJHSME/AMC 8 Problems and Solutions |
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