Product of two one-digit numbers greater than 5

This is a rule that helps remember a big part of the multiplication table. Assume you forgot the product 7·9. Do this. First find the excess of each of the multiples over 5: it's 2 for 7 (7 - 5 = 2) and 4 for 9 (9 - 5 = 4). Add them up to get 6 = 2 + 4. Now find the complements of these two numbers to 5: it's 3 for 2 (5 - 2 = 3) and 1 for 4 (5 - 4 = 1). Remember their product 3 = 3·1. Lastly, combine thus obtained two numbers (6 and 3) as 63 = 6·10 + 3.

The explanation comes from the following formula:

(5 + a)(5 + b) = 10(a + b) + (5 - a)(5 - b)

In our example, a = 2 and b = 4.

Another example:

 8×9= (5 + 3)(5 + 4)
  = 10(3 + 4) + (5 - 3)(5 - 4)
  = 70 + 2×1
  = 72.

Related material
Read more...

  • Multiplication by 9, 99, 999, (Multiply + Subtract) etc.
  • Squaring 2-Digit Numbers
  • Division by 5
  • Multiplication by 2
  • Multiplication by 5
  • Multiplication by 9, 99, 999, etc. (Something Special)
  • Product of 10a + b and 10a + c where b + c = 10
  • Product of numbers close to 100
  • Product of 2-digit numbers
  • Squaring Numbers in Range 26-50
  • Squaring Numbers in Range 51-100
  • Squares of Numbers That End in 5
  • Squares Can Be Computed Squentially
  • How to Compute Fast Any Square
  • Adding a Long List of Numbers

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