There are many things that are cool about divisibility rules, but this is really cool:

Let's take the multiples of 7, for example.

A cool thing about multiples of 7 is that if you multiply the last digit by 5, and add it to the remaining number, it will always be a multiple of 7. For example:

*84*: 8 + \(4 \times 5\) = 28 = \(4 \times 7\)

*161*: 16 + \(1 \times 5\) = 21 = \(7 \times 3\)

And this does not only apply to multiples of 7. Take multiples of 13, for example. When the last digit is multiplied by 4, the sum of that and the remaining number will also be a multiple of 13. For example:

*26*: 2 + \(6 \times 4\) = 26 (this kind of becomes a loop, but you get the point).

I have documented how this works for many numbers… (From what I have seen, it only occurs with some numbers)

*3*: Multiply last digit by **1** (3 * 3 = **1**0 − 1)

*7*: Multiply last digit by **5** (7 * 7 = **5**0 − 1

*9*: Multiply last digit by **1** (9 * 1 = **1**0 − 1)

*11*: Multiply last digit by **10** (11 * 9 = **10**0 − 1)

*13*: Multiply last digit by **4** (13 * 3 = **4**0 − 1)

*17*: Multiply last digit by **12** (17 * 7 = **12**0 − 1)

*19*: Multiply last digit by **2** (19 * 1 = **2**0 − 1)

The notes in parentheses are observations I have made. But can someone explain why this is all true????

## Comments

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TopNewestDoes this work for \(p=2, 5 \)? – Calvin Lin Staff · 1 year, 11 months ago

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– Ashwin Padaki · 1 year, 11 months ago

Well, this is just an idea, but I'm thinking it is related to the fact that no multiples of 2 or 5 can possibly have a units digit of 9.Log in to reply