当一个整数 a 不能被正整数 b 整除时，做除法后得到的"余数" r 满足：
When integer a is not divisible by positive integer b, the remainder r satisfies:
关键结论：
- 余数一定小于除数：r < b The remainder is always less than the divisor
- 余数可以为零（此时就是整除）Remainder can be zero (meaning divisible)
- 任何整数除以 b 的余数只有 b 种可能：0, 1, 2, ..., b−1 There are exactly b possible remainders
举例：
- 17 ÷ 5 = 3 余 2 → 17 = 5 × 3 + 2
- 23 ÷ 7 = 3 余 2 → 23 = 7 × 3 + 2
- 100 ÷ 3 = 33 余 1 → 100 = 3 × 33 + 1
一个数除以 7 余 3，除以 5 余 2。这个数最小是多少？ A number leaves a remainder of 3 when divided by 7, and a remainder of 2 when divided by 5. What is the smallest such number?
从中找"除以5余2"的：
3 ÷ 5 = 0 余 3 ✗
10 ÷ 5 = 2 余 0 ✗
17 ÷ 5 = 3 余 2 ✓
所以最小的数是 17。 Numbers ≡ 3 (mod 7): 3, 10, 17, 24... Check each mod 5: 17 ÷ 5 = 3 remainder 2 ✓. Answer: 17.
余数有以下重要的运算性质，先分别取余，再运算，再取余：
Remainders have the following arithmetic properties: reduce first, then operate, then take remainder.
| 运算 | 规则 | 说明 |
|---|---|---|
| 加法 | (a+b) mod n = [(a mod n) + (b mod n)] mod n | 先分别取余，再加，再取余 |
| 减法 | (a−b) mod n = [(a mod n) − (b mod n)] mod n | 先分别取余，再减，再取余 |
| 乘法 | (a×b) mod n = [(a mod n) × (b mod n)] mod n | 先分别取余，再乘，再取余 |
| 乘方 | ap mod n = (a mod n)p mod n | 底数先取余，再乘方，再取余 |
一个正整数 N 除以 3 余 2，除以 4 余 1，除以 5 余 3。满足条件的最小正整数 N 是多少？ A positive integer N leaves remainders of 2, 1, and 3 when divided by 3, 4, and 5 respectively. What is the smallest such N?
从中找同时满足另外两个条件的：
检验 53：
53 ÷ 3 = 17 余 2 ✓
53 ÷ 4 = 13 余 1 ✓
53 ÷ 5 = 10 余 3 ✓
三个条件全部满足！答案为 53。 Numbers ≡ 3 (mod 5): 3, 8, 13,..., 53. Check 53: 53÷3=17r2✓, 53÷4=13r1✓, 53÷5=10r3✓. Answer: 53.
"a mod n" 表示 a 除以 n 的余数。模运算在 AMC 中最常见的应用是寻找周期规律。
"a mod n" denotes the remainder when a is divided by n. The most common application in AMC is finding periodic patterns.
例：观察 2 的乘方除以 7 的余数规律
Example: Observe the pattern of remainders of powers of 2 when divided by 7
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | ... |
|---|---|---|---|---|---|---|---|---|
| 2n mod 7 | 2 | 4 | 1 | 2 | 4 | 1 | 2 | ... |
余数呈现 2, 4, 1 的周期循环，周期为 3。利用周期性，我们可以快速计算很大的乘方除以某数的余数。
The remainders follow a repeating pattern 2, 4, 1 with period 3. Using periodicity, we can quickly find remainders of very large powers.
② 列出前几项余数，找到周期
③ 用指数除以周期得到余数，确定答案
② List the first few remainders to find the period
③ Divide the exponent by the period to locate the answer
22024 除以 7 的余数是多少？ What is the remainder when 22024 is divided by 7?
2¹ mod 7 = 2，2² mod 7 = 4，2³ mod 7 = 1
2⁴ mod 7 = 2，2⁵ mod 7 = 4，2⁶ mod 7 = 1
余数呈 2, 4, 1 的周期，周期 = 3。
用 2024 ÷ 3 = 674 余 2。
余数为 2 对应周期中的第 2 项，即 4。 Powers of 2 mod 7: 2, 4, 1 (period 3). Since 2024 ÷ 3 = 674 remainder 2, the answer is the 2nd value = 4.