6.2 Fast Multiplication Using Distributivity

Once you see multiplication “by parts,” you can multiply some numbers in a single line, faster than a calculator, in your head.

Multiplying by 11

The trick rests on 11 = 10 + 1. For a number with digits d\,c\,b\,a (thousands, hundreds, tens, units), \overline{dcba} \times 11 = \overline{dcba}\times(10+1) = \overline{dcba0} + \overline{dcba}. Lining up the addition, each digit of the answer is the sum of two neighbouring digits of the original (carrying when a sum exceeds 9): \overline{dcba}\times 11 \;\rightarrow\; d \;\; (c+d) \;\; (b+c) \;\; (a+b) \;\; a.

Worked Example

2748 \times 11. Lay out the digits, then drop in the sum of each neighbouring pair (right to left, carrying when a sum tops 9):

  • units: 8
  • 4+8 = 12 → write 2, carry 1
  • 7+4 = 11,\ +1 = 12 → write 2, carry 1
  • 2+7 = 9,\ +1 = 10 → write 0, carry 1
  • thousands: 2 + 1 = 3

Reading off: \;2748 \times 11 = 30228.

Math Talk Evaluate using the rule: (i) 72\times11   (ii) 363\times11   (iii) 5184\times11   (iv) 3625471\times11.

Answers: (i) 792   (ii) 3993   (iii) 57024   (iv) 39880181. (For (iii): units 4; 8+4=122 carry 1; 1+8=9,+1=100 carry 1; 5+1=6,+1=7; thousands 5. Result 57024.)

Multiplying by 101, 1001, 10001, …

Now 101 = 100 + 1, so \overline{dcba}\times 101 = \overline{dcba00} + \overline{dcba}. The two copies overlap by two places, so we add digits that are two apart: \overline{dcba}\times 101 \;\rightarrow\; d\;\; c\;\; (b+d)\;\; (a+c)\;\; b\;\; a. For 1001 = 1000+1 the copies overlap by three places, for 10001 by four, and so on.

Worked Example

418267 \times 1001. Since 1001 = 1000+1: 418267 \times 1001 = 418267000 + 418267 = 418685267.

None of this is a modern invention. Indian scholars, Brahmagupta (628 CE), Śrīdharācārya (around 750 CE) and later Bhāskarācārya in the Līlāvatī (1150 CE), described shortcuts of exactly this kind, all powered by distributivity. In a follow-up verse (12.56) Brahmagupta gave such quick-multiplication recipes the name iṣṭa-guṇana.

Math Talk Split each second factor into a near round number plus or minus a little: (i) 76\times101   (ii) 838\times101   (iii) 372914\times1001   (iv) 2222\times1001   (v) 8657\times99   (vi) 41263\times999.

When the multiplier is 99 or 999, lean on subtraction instead: 99 = 100-1, 999 = 1000 - 1.

Answers: (i) 7676   (ii) 84638   (iii) 373286914   (iv) 2224222   (v) 8657(100-1) = 865700 - 8657 = 857043   (vi) 41263(1000-1) = 41263000 - 41263 = 41221737.