The 1089 Trick: Reverse, Subtract, Reverse and Add | 1089 数字谜题：倒序、相减、再倒序相加

📚 The 1089 Trick: Reverse, Subtract, Reverse and Add | 1089 数字谜题：倒序、相减、再倒序相加

The 1089 trick is one of the most surprising number puzzles in the IGCSE Mathematics syllabus. It looks like magic, but it can be explained fully using place value and basic algebra. Once you learn the proof, you will never see three-digit numbers in the same way again.

1089 数字谜题是 IGCSE 数学大纲中最令人惊讶的数字游戏之一。它看起来像魔术，但完全可以用位值和基础代数来解释。一旦你学会证明，你就会对三位数产生全新的认识。

1. What Is the 1089 Trick? | 什么是 1089 数字谜题？

Choose any three-digit number whose hundreds digit and units digit differ by at least 2. Write the number, reverse its digits, subtract the smaller number from the larger one, reverse the result, and then add this reversed result to the previous difference. The final answer is always 1089.

任选一个三位数，要求百位数字与个位数字相差至少 2。写出这个数，将数字倒序，用较大的数减去较小的数，再把所得结果倒序，最后把倒序后的数与刚才的差相加。最终答案总是 1089。

The rule works for numbers such as 321, 584, 903, and 476. It does not depend on the middle digit, and it does not depend on how large the difference between the first and last digits is, as long as that difference is at least 2.

这个规则适用于 321、584、903 和 476 等数。它与十位数字无关，也不取决于百位和个位相差多大，只要差值至少为 2 即可。

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2. A Worked Example | 例题演示

Take the number 521. Its reverse is 125. Because 521 is larger than 125, subtract 125 from 521 to get 396. Then reverse 396 to get 693. Finally add 396 and 693 to obtain 1089.

以数字 521 为例。它的倒序是 125。因为 521 大于 125，所以用 521 减 125，得到 396。然后把 396 倒序，得到 693。最后将 396 和 693 相加，得到 1089。

Step	Calculation	中文
Original number	521	原数
Reversed number	125	倒序数
Subtract	521 − 125 = 396	相减
Reverse the result	693	倒序结果
Add	396 + 693 = 1089	相加

742 to get 247. Subtract: 742 − 247 = 495. Reverse 495 to get 594. Add: 495 + 594 = 1089.

再试一个数字：742。将 742 倒序得到 247。相减：742 − 247 = 495。把 495 倒序得到 594。相加：495 + 594 = 1089。

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3. The Place-Value Setup | 位值法设定

Let the original three-digit number be written as 100a + 10b + c, where a, b and c are digits. Here a is the hundreds digit, b is the tens digit, and c is the units digit. We need a to be non-zero, and we assume a is greater than c.

设原始三位数为 100a + 10b + c，其中 a、b、c 是数字。这里 a 是百位数字，b 是十位数字，c 是个位数字。我们需要 a 不为零，并假设 a 大于 c。

N = 100a + 10b + c, M = 100c + 10b + a

The reversed number is 100c + 10b + a. If we subtract the smaller number from the larger one, we subtract M from N. The tens digit b cancels out completely.

倒序后的数是 100c + 10b + a。如果用较大的数减去较小的数，也就是用 N 减去 M。十位数字 b 会完全抵消。

N − M = (100a + 10b + c) − (100c + 10b + a) = 99(a − c)

This shows that the difference is always a multiple of 99. The value depends only on the difference between the hundreds digit and the units digit, not on the actual digits themselves.

这表明差值总是 99 的倍数。这个值只取决于百位数字与个位数字的差，而与具体数字本身无关。

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4. Algebraic Proof | 代数证明

ick to keep a three-digit difference, we require d ≥ 2, so d is between 2 and 9 inclusive.

设 d = a − c。因为 a 和 c 是数字且 a 大于 c，所以 d 可以是 1、2、3、4、5、6、7、8 或 9。为了使标准的谜题保持三位数差，我们要求 d ≥ 2，因此 d 在 2 到 9 之间。

The difference is 99d. We can rewrite 99d in expanded form:

差值是 99d。我们可以把 99d 写成展开形式：

99d = 100(d − 1) + 90 + (10 − d)

d. For example, if d = 2, the difference is 198; if d = 3, it is 297; if d = 4, it is 396.

这意味着三位数差的各位数字分别是 d − 1、9 和 10 − d。例如，如果 d = 2，差为 198；如果 d = 3，差为 297；如果 d = 4，差为 396。

Now reverse this three-digit difference. The reversed number is:

现在将这个三位数差倒序。倒序后的数是：

Reverse = 100(10 − d) + 90 + (d − 1)

Add the difference and its reverse:

把差值和它的倒序数相加：

99d + Reverse = 100(d − 1 + 10 − d) + 180 + (10 − d + d − 1) = 900 + 180 + 9 = 1089

This algebra proves that the answer is always 1089, no matter which valid three-digit number you choose.

这个代数过程证明，无论你选择哪个符合条件的三位数，答案总是 1089。

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5. Why the Result Is Independent of the Middle Digit | 为什么结果与十位数字无关

In the subtraction step, the tens digit b appears in both the original number and the reversed number as 10b. When you subtract, 10b − 10b becomes 0, so b vanishes from the difference immediately.

在减法步骤中，十位数字 b 同时以 10b 的形式出现在原数和倒序数中。当你相减时，10b − 10b 变成 0，所以 b 立刻从差值中消失。

198. This is because they all have a − c = 2, even though their tens digits are different.

例如，321、351、371 和 391 的第一个差值都是 198。这是因为它们的 a − c 都等于 2，尽管十位数字不同。

This independence is a strong example of algebraic cancellation. It helps explain why the puzzle feels so magical: the middle digit seems unimportant, yet the number still works.

这种无关性是代数抵消的典型例子。它说明了为什么这个谜题感觉如此神奇：十位数字似乎不重要，但数字仍然有效。

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6. The Role of the Digit Difference | 百位与个位差值的作用

he first difference, but it does not change the final total. The table below shows all possible standard differences and their reverses.

d = a − c 的值控制第一个差值的大小，但不会改变最终总和。下表列出了所有可能的标准差值及其倒序数。

d	99d	Reverse of 99d	Sum
2	198	891	1089
3	297	792	1089
4	396	693	1089
5	495	594	1089
6	594	495	1089
7	693	396	1089
8	792	297	1089
9	891	198	1089

Notice that the hundreds digit and units digit of each difference always add to 9, and the tens digit is always 9. This structure guarantees the final result of 1089.

注意每个差值中的百位数字和个位数字相加总是等于 9，十位数字也总是 9。这种结构保证了最终结果是 1089。

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7. Special Cases and Common Mistakes | 特殊情况与常见错误

k work, you must write the difference as 099, reverse it to 990, and add 099 + 990 = 1089.

如果 a − c = 1，差值是 99，这只是一个两位数。许多学生会把 99 倒序成 99，然后计算 99 + 99 = 198，这是错误的。要使谜题成立，你必须把差值写成 099，倒序成 990，再计算 099 + 990 = 1089。

ves 0 and the process stops. To avoid confusion, the standard rule states that the difference must be at least 2.

最常见的错误是选择百位和个位相差小于 2 的数。像 121 这样的数满足 a − c = 0，所以相减得到 0，过程就此结束。为了避免混淆，标准规则规定差值必须至少为 2。

Another mistake is reversing the result too early or forgetting to keep the digits aligned. Always write the difference as a three-digit number, even if it starts with zero, before reversing it.

另一个错误是过早倒序结果，或忘记保持数字对齐。倒序前，一定要把差值写成三位数，即使它以零开头。

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8. IGCSE Exam Links | IGCSE 考点链接

The 1089 trick connects several IGCSE Mathematics skills. It uses place value, subtraction with borrowing, reversing digits, expanding brackets, simplifying algebraic expressions, and constructing a logical proof. These are all tested in Cambridge IGCSE 0580 and 0980 papers.

1089 数字谜题连接了多个 IGCSE 数学技能。它涉及位值、借位减法、数字倒序、去括号、化简代数表达式以及构造逻辑证明。这些都是剑桥 IGCSE 0580 和 0980 试卷中考查的内容。

In exam questions, you may be asked to prove a number property or to show that an expression is always a multiple of a given integer. The method used here is exactly the type of reasoning examiners expect.

在考试题中，你可能会被要求证明某个数字性质，或证明某个表达式始终是某个整数的倍数。这里使用的方法正是考官期望的推理类型。

Practising this puzzle also helps with algebraic notation. You learn how a three-digit number can be expressed as 100a + 10b + c, which is a very common structure in IGCSE number and algebra questions.

练习这个谜题还有助于掌握代数记号。你将学会如何把三位数表示为 100a + 10b + c，这是 IGCSE 数与代数题中非常常见的结构。

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9. Practice Questions | 练习题

Q1. Use the 1089 trick with the number 804. Show each step clearly.

Q1. 用数字 804

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