AMC12 2003 B

AMC12 2003 B · Q7

AMC12 2003 B · Q7. It mainly tests Systems of equations, Money / coins.

Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is \$8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?

身无分文的皮特的小猪存钱罐里没有一分钱，但有 100 个硬币，全是镍币、角币和25美分币，总价值为 $8.35。它不一定包含所有三种类型的硬币。存钱罐里可能存在的角币数量的最大值和最小值之差是多少？

(A) 0 0

(B) 13 13

(D) 64 64

(E) 83 83

Answer

Correct choice: (D)

正确答案：（D）

Solution

Where $a,b,c$ is the number of nickels, dimes, and quarters, respectively, we can set up two equations: \[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\] Eliminate $a$ by subtracting $(2)$ from $(1)/5$ to get $b+4c=67$. Of the integer solutions $(b,c)$ to this equation, the number of dimes $b$ is least in $(3,16)$ and greatest in $(67,0)$, yielding a difference of $67-3=\boxed{\textbf{(D)}\ 64}$.

设 $a,b,c$ 分别为镍币、角币和25美分币的枚数，则有 \[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\] 将 $(1)$ 除以 $5$ 后减去 $(2)$ 以消去 $a$，得 $b+4c=67$。在该方程的整数解 $(b,c)$ 中，角币数 $b$ 的最小值出现在 $(3,16)$，最大值出现在 $(67,0)$，差为 $67-3=\boxed{\textbf{(D)}\ 64}$。

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