1958 AMC 12 第 45 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

45.

一张支票的金额为 xx 美元 yy 美分,其中 xxyy 都是两位数。误兑为 yy 美元 xx 美分,错误金额比正确金额多 $17.82\$17.82。则:

A check is written for xx dollars and yy cents, xx and yy both two-digit numbers. In error it is cashed for yy dollars and xx cents, the incorrect amount exceeding the correct amount by $17.82.\$17.82. Then:

xx 不可能大于 7070

xx cannot exceed 7070

yy 可以等于 2x2x

yy can equal 2x2x

支票金额不可能是 55 的倍数

the amount of the check cannot be a multiple of 55

错误金额可以等于正确金额的两倍

the incorrect amount can equal twice the correct amount

正确金额各位数字之和能被 99 整除

the sum of the digits of the correct amount is divisible by 99

答案:B
知识点:钱币数字一次方程
难度评级:1790
小提示:

将正确金额和错误金额都写成美分

Write the correct and incorrect amounts in cents

大提示:

两者之差化简为 99(yx)99(y-x)

Their difference simplifies to 99(yx)99(y-x)

解答:

以美分计,错误金额减正确金额为 (100y+x)(100x+y)=99(yx) \begin{aligned} &(100y+x)-(100x+y)\\ &\qquad=99(y-x) \end{aligned}\text{。}由于 $17.82\$17.82 等于 17821782 美分,99(yx)=1782 99(y-x)=1782\text{,}所以 yx=18y-x=18。两位数 x=18, y=36x=18,\ y=36 满足此关系,并且 y=2xy=2x。因此这种相等关系可以发生。

所以正确答案为 B

In cents, the incorrect amount minus the correct amount is (100y+x)(100x+y)=99(yx). \begin{aligned} &(100y+x)-(100x+y)\\ &\qquad=99(y-x). \end{aligned} Since $17.82\$17.82 is 17821782 cents, 99(yx)=1782, 99(y-x)=1782, so yx=18.y-x=18. The two-digit values x=18, y=36x=18,\ y=36 satisfy this relation and have y=2x.y=2x. Thus that equality can occur.

Therefore, the correct answer is B.

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