2015 AMC 12B 第 3 题

先试着解答 2015 AMC 12B 第 3 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 2015 AMC 12B 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。

3.

Isaac 写下了某个整数两次,另一个整数三次。这五个数的和为 100100,并且其中一个数是 2828。另一个数是多少?
Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100,100, and one of the numbers is 28.28. What is the other number?
88
1111
1414
1515
1818
答案:A
知识点:丢番图方程整除性
小提示:
设 xx 出现两次,yy 出现三次,则 2x+3y=1002x + 3y = 100。
Let xx appear twice and yy appear three times, so 2x+3y=1002x + 3y = 100
大提示:
分别检验 2828 是被写了两次还是三次,并保留整数情形
Test whether 2828 is the number written two times or three times, and keep the integer case
解答:
写作 2x+3y=1002x + 3y = 100。如果 2828 被写了两次,则 3y=100−56=443y = 100 - 56 = 44,这不是 33 的倍数。
所以 2828 被写了三次:2x=100−84=162x = 100 - 84 = 16,得 x=8x = 8。
因此,正确选项是 A。
Write 2x+3y=100.2x + 3y = 100. If 2828 were written twice, then 3y=100−56=44,3y = 100 - 56 = 44, which is not a multiple of 3.3.
So 2828 is written three times: 2x=100−84=16,2x = 100 - 84 = 16, giving x=8.x = 8.
Thus, the correct answer is A.
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