Casting out 9s

Most of us know the rule to check whether a number is divisible by 9. You sum the digits of the number and if the result is a multiple of 9, then the number itself is a multiple of 9. For example, 4572 is a multiple of 9 as 4 + 5 + 7 + 2 = 18 (or if you want to go further, 1 + 8 = 9) which is a multiple of 9. But not many of us know why this is so. Why should the numbers sum to a multiple of 9 (or to 9 if you keep summing till you end up with one digit)?

Well, it is because we have a decimal (base 10) number system. Take any general number, say 3587. We have 3587 = 3×1000 + 5×100 + 8×10 + 7×1, and each of the elements in the set {1, 10, 100, 1000,…} is one away from a multiple of 9 {0, 9, 99, 999,…}. This means 3587 is 5 (3 + 5 + 8 + 7 = 23, 2 + 3 = 5) away from a multiple of 9. This in essence is modulo 9 arithmetic; summing the digits gives us the remainder when a number is divided by 9. The same trick works for 3 insofar as checking for divisibility. Casting out 9s is useful to check the correctness of your arithmetic (all 4 basic operations). More on that here.

Can you extend the same logic to check why the following rule holds for divisibility by 11? (A number is divisible by 11 only if you arrive at 0 or 11 while alternately adding and subtracting the digits of the number, eg. 25476 is divisible by 11 as 2 – 5 + 4 – 7 + 6 = 0)

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2 comments

  1. Nice post. Have been doing divisibility tests for ever but did not know this. I haven't read through all the posts of your blog, but a quick comment: Love the title of your blog.

  2. Thank you and thank you :). The title was an inspired by my class mates who have been teasing me with that for the last five years.

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