AMC 8 2000 Problem 20
AMC8 2000 年第 20 题
特殊数与不定方程★★★☆☆
You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $1.02 , with at least one coin of each type. How many dimes must you have?
你有 9 枚硬币：若干便士，五分镍币，一角硬币和 25 分硬币，总价值是 1.02 美元，并且每种硬币都至少有一枚。那么你肯定有多少枚一角硬币？
- A.1
- B.2
- C.3
- D.4
- E.5
正确答案 · Correct Answer
A
解析 · Solution
Solution 1
Since you have one coin of each type, 1 + 5 + 10 + 25 = 41 cents are already determined, leaving you with a total of 102 - 41 = 61 cents remaining for 5 coins. You must have 1 more penny. If you had more than 1 penny, you must have at least 6 pennies to leave a multiple of 5 for the nickels, dimes, and quarters. But you only have 5 more coins to assign. Now you have 61 - 1 = 60 cents remaining for 4 coins, which may be nickels, quarters, or dimes. If you have only one more dime, that leaves 50 cents in 3 nickels or quarters, which is impossible. If you have two dimes, that leaves 40 cents for 2 nickels or quarters, which is again impossible. If you have three dimes, that leaves 30 cents for 1 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 60 cents in 4 coins must be divided between the quarters and nickels. We quickly see that 2 nickels and 2 quarters work. Thus, the total count is 2 quarters, 2 nickels, 1 penny, plus one more coin of each type that we originally subtracted. Double-checking, that gives a total 2 + 2 + 1 + 4 = 9 coins, and a total of cents. There is only 1 dime in that combo, so the answer is . 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
It is clear that there should only be 2 pennies; any more would take up too many coins, and the limit is 9. Now we have 1 left, and 7 coins to use. Looking at the quarters, we can make methodical guesses. If there is 1 quarter, then we would have to make 0.75 with 6 coins. We take a few educated guesses for the nickel and dime combos, and see that 1 quarter will not work. Trying values for 2 quarters, we see that this will not work either. When we reach 3 quarters, the remaining is 0.25 made from 4 coins. We try with 2 dimes, which does not work (it only takes 3 coins) and we try with 1 dime. After trying 2 pennies, 3 quarters, 1 dime, and 3 nickels, it is evident that this combo works. Therefore, the answer is It might seem that this solution takes time, but in reality, it can be done very quickly, given ones desire to solve a problem quickly
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