1955 AMC 12 第 37 题

先试着解答 1955 AMC 12 第 37 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1955 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

37.

一个三位数从左到右的数字为 hh、tt 和 uu,其中 h>uh>u。用原数减去各位数字倒序所得的数,差的个位数字为 44。从右到左接下来的两个数字是:
A three-digit number has, from left to right, the digits h,h, t,t, and uu with h>u.h>u. When the number with the digits reversed is subtracted from the original number, the units’ digit in the difference is 4.4. The next two digits, from right to left, are:
55 和 99
55 and 99
99 和 55
99 and 55
无法确定
impossible to tell
55 和 44
55 and 44
44 和 55
44 and 55
答案:B
知识点:digit reversal模运算subtraction
小提示:
以代数式相减:(100h+10t+u)(100h+10t+u) −(100u+10t+h)=99(h−u){}-(100u+10t+h)=99(h-u)
Subtract algebraically: (100h+10t+u)(100h+10t+u) −(100u+10t+h)=99(h−u){}-(100u+10t+h)=99(h-u)
大提示:
求使 99(h−u)99(h-u) 的个位数字为 44 的数位差 h−uh-u
Find the digit h−uh-u for which 99(h−u)99(h-u) ends in 44
解答:
差为 99(h−u)99(h-u)。由于 h−uh-u 是从 11 到 99 的整数,并且个位数字是 44,所以需要 9(h−u)≡4(mod10)9(h-u)\equiv4\pmod{10}。由此得 h−u=6h-u=6,因而差为 99⋅6=594。 99\cdot6=594\text{。}从右向左,在个位数字 44 之后的两个数字是 99 和 55。
因此,正确答案是 B。
The difference is 99(h−u).99(h-u). Since h−uh-u is an integer from 11 through 9,9, and the units digit is 4,4, we need 9(h−u)≡4(mod10).9(h-u)\equiv4\pmod{10}. This gives h−u=6,h-u=6, so the difference is 99⋅6=594. 99\cdot6=594. Moving from right to left after the units digit 4,4, the next digits are 99 and 5.5.
Thus, the correct answer is B.
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