AMC12 2025 B

AMC12 2025 B · Q23

AMC12 2025 B · Q23. It mainly tests Primes & prime factorization, GCD & LCM.

Let $S$ be the set of all integers $z > 1$ such that for all pairs of nonnegative integers $(x, y)$ with $x < y < z$, the remainder when $2025x$ is divided by $z$ is less than the remainder when $2025y$ is divided by $z$. What is the sum of the elements of $S$?

设$S$为所有整数$z > 1$的集合，使得对于所有非负整数对$(x, y)$满足$x < y < z$，$2025x$除以$z$的余数小于$2025y$除以$z$的余数。$S$的元素之和是多少？

(A) 3041 3041

(B) 3542 3542

(D) 4044 4044

(E) 4319 4319

Answer

Correct choice: (E)

正确答案：（E）

Solution

Notice that there are $z$ distinct remainders modulo $z$. However, if we let $R(x)$ denote the remainder when $2025x$ is divided by $z$, we know that by the problem condition, \[R(0)<R(1)<R(2)<\cdots<R(z-1)\] But there are only $z$ distinct remainders, and each of the $z$ terms above must correspond to a distinct remainder, so we must have $R(k)=k$ for all $k=0,1,\dots,z-1$. Then $2025k\equiv k\pmod{z}$, so $2024k\equiv 0\pmod{z}$. Since $k$ can vary, this is equivalent to $2024\equiv 0\pmod{z}$, or $z\mid 2024$. Therefore, the values $z$ that satisfy the property are the factors of $2024$, so we simply need to find the sum of the factors of $2024$ and subtract $1$ at the end since $z\ne 1$. But $2024=2^3\cdot 11\cdot 23$, so the sum of the factors minus $1$ is \[(1+2+4+8)(1+11)(1+23)-1=15\cdot12\cdot24-1=4320-1=\boxed{\textbf{(E)}~4319 }.\]

注意到模$z$有$z$个不同的余数。但是，如果令$R(x)$表示$2025x$除以$z$的余数，根据题目条件， \[R(0)<R(1)<R(2)<\cdots<R(z-1)\] 但只有$z$个不同的余数，上面的$z$项必须对应不同的余数，因此必须有$R(k)=k$对所有$k=0,1,\dots,z-1$成立。那么$2025k\equiv k\pmod{z}$，所以$2024k\equiv 0\pmod{z}$。由于$k$可变，这等价于$2024\equiv 0\pmod{z}$，即$z\mid 2024$。因此，满足性质的$z$是$2024$的因数，所以只需找到$2024$的因数和并减去$1$（因为$z\ne 1$）。但$2024=2^3\cdot 11\cdot 23$，因数和减$1$为 \[(1+2+4+8)(1+11)(1+23)-1=15\cdot12\cdot24-1=4320-1=\boxed{\textbf{(E)}~4319 }\]。

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