AMC12 2021 B

AMC12 2021 B · Q10

AMC12 2021 B · Q10. It mainly tests Linear equations, Casework.

Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?

从集合$\{1,2,3,4,\dots,36,37\}$中选出两个不同的数，使得剩余$35$个数的和等于这两个数的乘积。这两个数的差是多少？

(A) 5 5

(B) 7 7

(D) 9 9

(E) 10 10

Answer

Correct choice: (E)

正确答案：（E）

Solution

The sum of the first $n$ integers is given by $\frac{n(n+1)}{2}$, so $\frac{37(37+1)}{2}=703$. Therefore, $703-x-y=xy$ Rearranging, $xy+x+y=703$. We can factor this equation by SFFT to get $(x+1)(y+1)=704$ Looking at the possible divisors of $704 = 2^6\cdot11$, $22$ and $32$ are within the constraints of $0 < x \leq y \leq 37$ so we try those: $(x+1)(y+1) = 22\cdot32$ $x+1=22, y+1 = 32$ $x = 21, y = 31$ Therefore, the difference $y-x=31-21=\boxed{\textbf{(E) }10}$.

前$n$个整数的和为$\frac{n(n+1)}{2}$，因此$\frac{37(37+1)}{2}=703$。因此，$703-x-y=xy$ 重排，$xy+x+y=703$。通过SFFT技巧因式分解得到 $(x+1)(y+1)=704$ 查看$704 = 2^6\cdot11$的因子对，$22$和$32$满足$0 < x \leq y \leq 37$的约束，因此尝试： $(x+1)(y+1) = 22\cdot32$ $x+1=22, y+1 = 32$ $x = 21, y = 31$ 因此，差$y-x=31-21=\boxed{\textbf{(E) }10}$。

Topics