4.1.3 Multiplying Two Numbers Whose Units Add to 10 and the Rest is the Same

This is a generalized version of the Squares Ending in 5 Trick (Section 1.2.8).

The Derivation

Take $n_1 = ab$ and $n_2 = ac$, where $b + c = 10$:

$$ab \times ac = (10a + b)(10a + c) = 100a(a + 1) + bc$$

The Rule

For two numbers with the same leading digit(s) and units that add to 10:

1. Last two digits = units digits multiplied together
2. Remaining digits = leading digit(s) × (leading digit(s) + 1)

Note: The Squares Ending in 5 trick is a special case where $bc = 5 \times 5 = 25$.

Example: 68 × 62

Step	Calculation	Result
Tens/Ones	8 × 2	16
Remaining	6 × (6 + 1) = 6 × 7	42

Answer: 4216

Example: 173 × 177

Step	Calculation	Result
Tens/Ones	3 × 7	21
Remaining	17 × (17 + 1) = 17 × 18	306

Answer: 30621

Tip: You could also use “Multiplying Two Numbers Equidistant from a Third” (Section 1.2.10): 68 × 62 = 65² − 3² = 4225 − 9 = 4216. But this “new” trick cuts down on the subtraction step!

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Problem Set 4.1.3

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