魔方背后和群论有什么关系？
How is a Rubik's cube related to group theory?
每次转动都是一个置换，所有转动构成群，目标是回到单位元（复原）。群论能算出任意状态最少转动步数（上帝之数约20）。
Each turn is a permutation, and all turns form a group whose identity is the solved state. Group theory yields the maximum solution length, God's number, about 20 moves.
难度 专业 · 适合 高中(15-18)
CC BY 4.0 · 来源待核验 · Source pending review
来源 · Source
出处备注：本题原挂出处与答案内容不符，已撤回；具体书目待补，暂不展示来源。
本题原挂出处与答案内容不符，已撤回；具体书目待补。
Previous citation withdrawn as unsupported for this answer; a specific source is pending.
#趣味数学 #群论
照镜子时，为什么左右颠倒却不会上下颠倒？
In a mirror, why are left and right swapped but not up and down?
镜子其实没有左右颠倒，它沿前后方向反转。因为我们习惯转身面对面，把前后反转误读成了左右互换；上下轴没有被这样重映射。
The mirror does not flip left-right; it reverses front-back. We imagine turning to face the reflection, misreading that as a left-right swap, while the vertical axis is not remapped the same way.
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
Martin Gardner,《My Best Mathematical and Logic Puzzles》Dover,1994
M. Gardner, My Best Mathematical and Logic Puzzles, Dover, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL460047W：My best mathematical and
#趣味数学 #镜像
数字黑洞6174是怎么回事？
What is the 6174 number black hole?
任取一个四位不全相同的数，把数字按降序和升序各排一次相减，重复下去，最多七步必得到6174，之后不再变。这是卡普雷卡尔常数。
Take any 4-digit number with not all equal digits, form the descending and ascending numbers, subtract, and repeat. Within seven steps you always reach 6174, the Kaprekar constant, which then stays fixed.
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
Martin Gardner,《My Best Mathematical and Logic Puzzles》Dover,1994
M. Gardner, My Best Mathematical and Logic Puzzles, Dover, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL460047W：My best mathematical and
#趣味数学 #数字黑洞
用二进制怎么能“读心”猜数字？
How can binary representation 'read your mind' to guess a number?
把 1–63 分成 6 张表，每张表对应一个二进制位（1,2,4,8,16,32），只含该位为 1 的数。观众指出含其数字的表，把这些表的首数相加就是原数字——这是二进制的魔术。
Split 1–63 into 6 cards, each holding numbers with a given binary bit set (1,2,4,8,16,32). The sum of the first numbers of the cards a spectator points to reveals their number—binary magic.
难度 进阶 · 适合 初中(12-15)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Martin Gardner,《My Best Mathematical and Logic Puzzles》Dover,1994
M. Gardner, My Best Mathematical and Logic Puzzles, Dover, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL460047W：My best mathematical and
#数学魔术 #二进制
九点问题：如何用四条直线一笔连起九个点且不离开边界？
The nine-dot puzzle: connect nine dots with four straight lines without lifting the pen or leaving the grid?
关键是让直线超出九点的正方形边界。把线画到外面再折回，就能用四条线一笔连完，它常用来说明要打破思维框框。
The trick is to let the lines extend beyond the implied square of dots. Going outside and back lets you cover all nine with four lines, a classic lesson in thinking outside the box.
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
Martin Gardner,《My Best Mathematical and Logic Puzzles》Dover,1994
M. Gardner, My Best Mathematical and Logic Puzzles, Dover, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL460047W：My best mathematical and
#趣味数学 #思维定势
数独背后的数学原理是什么？
What is the math behind Sudoku?
数独是一个9×9的拉丁方：每行、每列和每个九宫格都恰好含1到9各一次。它属于组合设计里的拉丁方与约束满足问题。
Sudoku is a 9×9 Latin square where each row, column and box contains 1–9 exactly once. It belongs to combinatorics as a Latin square and a constraint-satisfaction problem.
难度 入门 · 适合 大众
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
《组合数学》Brualdi
Introductory Combinatorics by Richard Brualdi
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787111377870（豆瓣书目条目: 组合数学）
#趣味数学 #拉丁方
什么是幻方？
What is a magic square?
幻方是把数字排成方阵，使每行、每列及两条对角线的和都相等。最著名的是洛书（三阶幻方）：每行每列和为15。它兼具数学与游戏之美。
A magic square arranges numbers so every row, column and diagonal sums equally; the classic 3×3 Lo Shu square sums to 15 throughout.
难度 入门 · 适合 小学低段(6-8)
CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
来源 · Source
Martin Gardner,《My Best Mathematical and Logic Puzzles》Dover,1994
M. Gardner, My Best Mathematical and Logic Puzzles, Dover, 1994
独立核验依据 · Independent verification: 核验依据 2 OPENLIBRARY 标识 /works/OL460047W：My best mathematical and
#数学魔术 #幻方
鸡兔同笼：头35、脚94，鸡兔各几只？
Chicken-rabbit cage: 35 heads and 94 feet, how many of each?
假设全是鸡则有70只脚，多出24只脚，每只兔比鸡多2脚，所以兔有12只，鸡有23只。这是用假设法解二元一次问题。
If all were chickens there would be 70 feet; the extra 24 feet mean 12 rabbits (each has 2 more feet), leaving 23 chickens. It is a guess-and-check way to solve two equations.
难度 入门 · 适合 小学低段(6-8)
CC BY 4.0 · 已核验 · Verified 2026-11-02
来源 · Source
《数学花园漫游记》,马希文,中国少年儿童出版社,2004
Ma Xiwen, Adventures in the Math Garden, China Children Press, 2004
独立核验依据 · Independent verification: 核验依据 1 ISBN 9787500767558（豆瓣书目条目: 数学花园漫游记）
#趣味数学 #鸡兔同笼