2021 AMC 12B Spring 第 10 题

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10.

从集合 {1,2,3,4,,36,37}\{1,2,3,4,\ldots,36,37\} 中选出两个不同的数,使剩下 3535 个数的和等于这两个数的乘积。 这两个数的差是多少?

Two distinct numbers are selected from the set {1,2,3,4,,36,37}\{1,2,3,4,\ldots,36,37\} so that the sum of the remaining 3535 numbers is the product of these two numbers. What is the difference of these two numbers?

55

77

88

99

1010

答案:E
知识点:西蒙最爱的因式分解技巧等差数列
难度评级:1530
解答:

总和为 1+2++37=7031+2+\cdots+37=703。若选出的两个数为 aabb,则 703ab=ab703-a-b=ab

所以 ab+a+b=703ab+a+b=703, 两边加 11(a+1)(b+1)=704=2611(a+1)(b+1)=704=2^6\cdot 11

我们需要 a+1,b+1a+1,b+1223838 之间的因数。因数对 2232=70422\cdot 32=704 可行,得到 a=21a=21b=31b=31

它们的差为 3121=1031-21=10

所以正确答案是 E

The sum 1+2++37=703.1+2+\cdots+37=703. If the chosen numbers are aa and b,b, then 703ab=ab.703-a-b=ab.

So ab+a+b=703,ab+a+b=703, and adding 11 gives (a+1)(b+1)=704=2611.(a+1)(b+1)=704=2^6\cdot 11.

We need factors a+1,b+1a+1,b+1 between 22 and 38.38. The pair 2232=70422\cdot 32=704 works, giving a=21,a=21, b=31.b=31.

Their difference is 3121=10.31-21=10.

Thus, the correct answer is E.

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