身份证最后一位校验码，是用什么数学算出来的？
What math produces the last check digit of an ID number?
前17位分别乘以固定权重后求和，再对11取模，余数对应一个校验字符（0到9和X）。这样能立刻发现输错或漏一位的错误。
The first 17 digits are weighted, summed, then taken modulo 11, mapping the remainder to a check character (0–9 or X). This instantly catches single-digit or missing-digit errors.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Graham, Knuth, Patashnik,《Concrete Mathematics》第2版,Addison-Wesley,1994
R. L. Graham et al., Concrete Mathematics, 2nd ed., Addison-Wesley, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL3951639W：Concrete mathematics（按标
#生活中的密码 #校验码
超市条形码里藏着什么简单的数学？
What simple math is hidden in a supermarket barcode?
条形码用宽窄条表示0和1的二进制，并带一位奇偶校验位来发现扫描错误。它把商品编号编码成机器能快速读取的数字串。
A barcode encodes binary 0/1 with bar widths and includes a parity check digit to catch scan errors. It turns a product code into a machine-readable number string.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Graham, Knuth, Patashnik,《Concrete Mathematics》第2版,Addison-Wesley,1994
R. L. Graham et al., Concrete Mathematics, 2nd ed., Addison-Wesley, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL3951639W：Concrete mathematics（按标
#生活中的密码 #条形码
为什么密码长一位，破解难度会大很多？
Why does adding one character to a password hugely raise cracking difficulty?
若每个位置有约62种字符，n位密码有62的n次方种组合，长度加一组合就乘62。指数增长让长密码在暴力破解面前极其安全。
With about 62 possible characters per position, an n-character password has 62^n combinations; one more character multiplies that by 62. Exponential growth makes long passwords very safe against brute force.
难度 进阶 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Graham, Knuth, Patashnik,《Concrete Mathematics》第2版,Addison-Wesley,1994
R. L. Graham et al., Concrete Mathematics, 2nd ed., Addison-Wesley, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL3951639W：Concrete mathematics（按标
#生活中的密码 #组合