Multiply Two Numbers Between 10 and 20 in Three Steps

Multiplying any two numbers between 10 and 20 takes three mental steps once you know the tens-plus-excess method.

This article derives the formula from scratch, works through four examples re-checked from first principles, covers what happens at the boundaries (10 and 20), and shows where this saves time in placement aptitude rounds.

The Three-Step Method

Every number in the 10-to-20 range is 10 plus a single digit. That digit is called the excess. For 14 the excess is 4. For 17 the excess is 7. For 20 the excess is 10.

Given two numbers a and b, both in [10, 20]:

• Step 1: Add a + b, then subtract 10. Call the result the tens multiplier.
• Step 2: Multiply the tens multiplier by 10.
• Step 3: Multiply the two excess digits, (a - 10) and (b - 10), then add the result to Step 2.

The complete formula is a × b = (a + b - 10) × 10 + (a - 10) × (b - 10).

A quick preview before the full examples. For 14 × 16:

• Step 1: 14 + 16 - 10 = 20
• Step 2: 20 × 10 = 200
• Step 3: 4 × 6 = 24
• Answer: 200 + 24 = 224

Compare this with the column method: multiply 14 by 6 (which requires carrying: 6 × 4 = 24, write 4 carry 2; 6 × 1 + 2 = 8, giving 84), then add 14 × 10 = 140 to get 224. The teens method sidesteps the carry entirely, because Step 3 always involves two single-digit numbers (or at most one digit that is 10, in the boundary case).

Why This Works: The Algebra

Write a = 10 + p and b = 10 + q, where p and q are the excess digits.

Expand the product:

(10 + p) × (10 + q) = 100 + 10p + 10q + pq

Factor the first three terms:

= 10 × (10 + p + q) + pq

Since 10 + p + q = a + b - 10:

= (a + b - 10) × 10 + (a - 10) × (b - 10)

This identity is exact. No rounding, no approximation. The trick works because 10 is the decimal base: multiplying by 10 is a single shift, and both inputs sit within 10 of that base. The same structural insight drives the shift-and-add method for multiplying any number by 111, where the number’s decimal digits play a similar role.

Four Verified Examples

All four re-derived from first principles; the column check confirms each answer independently.

Example 1: 14 × 16

• Excess digits: 14 - 10 = 4 and 16 - 10 = 6
• Step 1: 14 + 16 - 10 = 20
• Step 2: 20 × 10 = 200
• Step 3: 4 × 6 = 24
• Answer: 200 + 24 = 224
• Column check: 14 × 10 + 14 × 6 = 140 + 84 = 224 ✓

Example 2: 13 × 18

• Excess digits: 13 - 10 = 3 and 18 - 10 = 8
• Step 1: 13 + 18 - 10 = 21
• Step 2: 21 × 10 = 210
• Step 3: 3 × 8 = 24
• Answer: 210 + 24 = 234
• Column check: 13 × 20 - 13 × 2 = 260 - 26 = 234 ✓

Example 3: 17 × 19

• Excess digits: 17 - 10 = 7 and 19 - 10 = 9
• Step 1: 17 + 19 - 10 = 26
• Step 2: 26 × 10 = 260
• Step 3: 7 × 9 = 63
• Answer: 260 + 63 = 323
• Column check: 17 × 20 - 17 × 1 = 340 - 17 = 323 ✓

Example 4: 15 × 13

• Excess digits: 15 - 10 = 5 and 13 - 10 = 3
• Step 1: 15 + 13 - 10 = 18
• Step 2: 18 × 10 = 180
• Step 3: 5 × 3 = 15
• Answer: 180 + 15 = 195
• Column check: 15 × 10 + 15 × 3 = 150 + 45 = 195 ✓

All four answers verified. The column checks use a different decomposition in each case so the two methods are genuinely independent, not circular.

Edge Cases: When One Number Is 10 or 20

The formula works cleanly at the boundaries, but the behaviour is worth seeing before you meet it under exam conditions.

When one number is 10

Let a = 10, so the excess a - 10 = 0. Step 3 becomes 0 × (b - 10) = 0. The formula collapses to:

10 × b = (10 + b - 10) × 10 + 0 = b × 10

No trick is needed here. The result is immediate. Knowing that the formula degenerates correctly is a useful sanity check.

• Example: 10 × 17
• Step 1: 10 + 17 - 10 = 17
• Step 2: 17 × 10 = 170
• Step 3: 0 × 7 = 0
• Answer: 170 + 0 = 170 ✓

When one number is 20

Let b = 20, so the excess b - 10 = 10. Step 3 becomes (a - 10) × 10, which is always a round multiple of 10 and adds without carrying.

• Example: 20 × 14

• Step 1: 20 + 14 - 10 = 24

• Step 2: 24 × 10 = 240

• Step 3: 10 × 4 = 40

• Answer: 240 + 40 = 280 ✓

• Column check: 20 × 10 + 20 × 4 = 200 + 80 = 280 ✓

• Example: 20 × 20

• Step 1: 20 + 20 - 10 = 30

• Step 2: 30 × 10 = 300

• Step 3: 10 × 10 = 100

• Answer: 300 + 100 = 400 ✓

The only extra mental load when both numbers are near 20 is that Step 3 produces a three-digit number (100) rather than a two-digit one. The addition is still straightforward. The trick remains faster than a full column multiplication.

Where This Saves Time in Placement Tests

The quantitative aptitude sections of TCS NQT and AMCAT include multiplication-based questions where candidates must evaluate or verify a product under time pressure. eLitmus pH Test quant rounds have the same problem type at similar difficulty. In each of these tests, the questions that hit the 10-to-20 range appear alongside harder problems on ratios, percentages, and data interpretation, making fast multiplication a support skill rather than a terminal one.

The value is in the compound effect. A 30-question Numerical Ability section may include five or six problems where this trick applies directly. That means finding a product quickly, eliminating a wrong MCQ option by mental computation, or confirming a ratio. Saving 10 seconds on each of those six problems releases a full minute, which may cover the time needed for a harder problem at the end. That compound effect is also why the same derive-don’t-memorise approach works well for calendar problems in aptitude tests and for clocks aptitude questions, where the underlying rule is more useful than any memorised shortcut table.

The three-step format compresses a multi-digit multiplication into one addition and one small-number product. That systematic reduction (identifying the structure to exploit, then applying it consistently) is the same reasoning habit that matters when debugging an LLM prompt or tracing a failure in a retrieval pipeline. TinkerLLM is a self-paced playground at ₹299 where those first-principles habits are applied to actual model interactions, starting from a build problem rather than a lecture.