Hey there! Let's break this down step by step to prove why this trick works for any two-digit numbers, and then talk about how to extend it to larger numbers.

一、两位数乘法技巧的数学证明 #

First, let's formalize any two-digit numbers with variables to make this general. Let the first number be (10a + b) (where (a) is the tens digit, 1≤a≤9; (b) is the units digit, 0≤b≤9) and the second number be (10c + d) (similarly, 1≤c≤9, 0≤d≤9).

The actual product of these two numbers expands to:

(10a + b)(10c + d) = 100ac + 10ad + 10bc + bd

Now let's map this to the trick you described:

1. Step 1: Concatenate (a×c) and (b×d)
   When you concatenate these two results, you're actually creating the number (100ac + bd) (in your example, (7×9=63) and (8×4=32) concatenates to 6332, which is (63×100 + 32 = 6332)).
2. Step 2: Calculate (b×c + a×d), shift left by one, and add
   Shifting the sum left by one digit is the same as multiplying it by 10. So this part gives us (10×(ad + bc)) (your example: (8×9 +7×4=100), shifted left becomes 1000, which is (10×100)).

If we add these two parts together:

(100ac + bd) + 10(ad + bc) = 100ac + 10ad +10bc + bd

This is exactly the expanded form of our original product! That's why this trick works for any pair of two-digit numbers—it's just a rearranged version of the standard algebraic expansion, packaged as a mental math shortcut.

二、扩展到更大的数字 #

This trick is actually a simplified version of a broader method called cross multiplication (or lattice multiplication). For numbers with more digits, the core idea stays the same: multiply every digit of the first number by every digit of the second, sum the products that belong to the same place value (units, tens, hundreds, etc.), then handle carry-over as needed.

Let's use a 3-digit example to show how this works: multiply (123 × 456):

1. Label each digit by its place value:
   • First number: (1) (hundreds place), (2) (tens), (3) (units)
   • Second number: (4) (hundreds), (5) (tens), (6) (units)
2. Calculate all digit pairs and group them by the total place value (sum of their digit positions, starting from 0 for units):
   • Place value 4 (ten-thousands): (1×4 = 4)
   • Place value 3 (thousands): (1×5 + 2×4 = 5 + 8 =13)
   • Place value 2 (hundreds): (1×6 +2×5 +3×4=6+10+12=28)
   • Place value 1 (tens): (2×6 +3×5=12+15=27)
   • Place value 0 (units): (3×6=18)
3. Process from right to left, handling carry-over:
   • Units place: 18 → write down 8, carry over 1
   • Tens place: 27 +1=28 → write down 8, carry over 2
   • Hundreds place:28+2=30 → write down 0, carry over3
   • Thousands place:13+3=16 → write down6, carry over1
   • Ten-thousands place:4+1=5 → write down5
4. The final result is 56088, which matches the actual product (123×456=56088).

For even larger numbers (4-digit, 5-digit, etc.), you just repeat the pattern: multiply every digit pair, sum the products for each combined place value, then work through the carry-over from right to left.

备注：内容来源于stack exchange，提问作者SirMrpirateroberts