Product of 10a + b and 10a + c where b + c = 10

This is quite similar to the squaring of numbers that end with 5.

For example, compute 113×117, where a = 11, b = 3, and c = 7. First compute 11·(11 + 1) = 11·12 = 132 (since 3 = 1 + 2). Next, append 21 (= 3×7) to the right of 132 to get 13221!

Why does this work?

 (10a + b)(10a + c)= 100a² + 10a·(b + c) + bc
  = 100a(a + 1) + bc.

Another example: compute 242×248, with a = 24, b = 2, and c = 8. Then

24·(24 + 1) = 24² + 24 = 576 + 24 = 600.

(Also 24×25 = 25² - 25 = 625 - 25 = 600.)

Now appending 16 (= 2×8) we get 242×248 = 60016.


[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Algebra| |Rapid math|

 

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]