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So I'm just going through KhanAcademy to refresh my basic pre-arithmetic and although it's embarassing I thought I'd get this thing checked up just for safety:

https://www.khanacademy.org/math/arithmetic/factors-multiples/divisibility_tests/v/divisibility-tests-for-2--3--4--5--6--9--10

In this video we check whether we can divide large numbers into 2, 3, 4, 5, 6, 9, 10. For checking whether the number was divisible by 2 it was simply whether the last number was even or odd. For 3 it was adding up all the digits in the number and seeing whether that was divisible by 3. My question is, does this method work for every number? Just add all the digits up and see whether it's divisible by the number? He used other methods for the preceeding numbers but I carried on with the adding of all the digits method and it seemed to work for the few I tried.

I appreciate any help, thanks a lot.

EDIT: Upon further tests it seems apparent that it's not. My bad.

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2 Answers 2

nope it does not. For example to check divisibility by 11 check if the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11.

So is not alway the sum of the digits that gives you information on divisbility by a certain number.

For example check HERE for how to check if a number is divisible by 7...

Can get quite complicated.

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Could you not just divide it by the number and check whether it has a remainder? –  user2925800 Jan 26 at 15:45
    
Yeah of course. But usually those other methods are so easy that can be applied also to very big numbers. For example to check if a number is divisible by 2 you need just the last digit. The number could have 2000 digits but you don't actually need to do the division by 2! –  Umberto Jan 26 at 16:05
    
Thanks a lot! Just going through some of the rules now and they seem easy enough. –  user2925800 Jan 26 at 16:12

The method works for $3$ and for $9$. It clearly fails for $2$ ($12$ is divisible by $2$ but $1+2$ is not; $11$ is not divisible by $2$, but $1+1$ is). The fact that it works for $3$ and $9$ is based on the observation that $10$ and $1$ leave the same remainder upon division (i.e. the difference $10-1$ is a multiple of $3$ and of $9$).

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Thanks a lot. Do you know whether you could just divide the number by the number or is it important to learn all the induvidual rules for the numbers? –  user2925800 Jan 26 at 15:53

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