Multiples Rule for 7

Update: check out a new rule for divisibility by seven.

Last month I posted a bit about the rule to tell if a number is divisible by three or nine. With these rules in place, here are the rules we have to tell whether a number is divisible by any given digit:

1
it will be. every number is divisible by one

2
last digit is 0,2,4,6, or 8 (number is even)

3
sum of digits is a multiple of three
example
73901 -> 7+3+9+0+1 = 20 -> 2 + 0 = 2, so 73901 is not a multiple of 3

4
the following quantity must be even: half the ones digit plus the tens digit
examples
57042 -> \frac{1}{2} 2 + 4 = 5 is odd: 57042 is not a multiple of 4.
2236 -> \frac{1}{2} 6 + 3 = 6, 2236 is a multiple of 4.
901 -> \frac{1}{2} 1 + 0 = \frac{1}{2}: 901 is not a multiple of 4.

5
ends in zero or five

6
multiple of both 3 and 2. i.e. must be even and sum of digits must be a multiple of three

8
the following quantity must be even: half of (half of the ones digit plus the tens digit) plus hundreds digit
examples
690576 -> \frac{\frac{6}{2} + 7}{2}+5 = 10: 690376 is a multiple of 8.
532 -> \frac{\frac{2}{2}+3}{2}+5 = 7: 532 is not a multiple of 8

9
the sum of the digits must be a multiple of nine

You’ll notice I left off 7. There is a rule, although it’s a bit more complicated. Let’s just take an example, say 18,913. We can use a rule similar to the one for three and nine. The rule for three and nine worked because they evenly divide, 9, 99, 999, 9999 etc. 7 doesn’t do that, but we can find how close it comes. 7 goes into 7, 98, 994, and 9996.

18,913 = 1*10,000 + 8*1,000 + 9*100 + 1*10 + 3 = 1*(9996+4) + 8*(994+6) + 9*(98+2) + 1*(7+3) + 3

We want to know if that’s a multiple of seven, so divide it all by seven and see if you get an integer. Conveniently, the 9996/7, 994/7, etc are all integers. That means we only need to know if the following is a multiple of seven:
1*4 + 8*6 + 9*2 + 1*3 + 3*1.

This is a general pattern: a number is a multiple of seven if the ones digit plus three times the tens digit plus 2 times the hundreds digit plus six times the thousands digit plus 4 times the ten-thousands digit is a multiple of seven.

But what if we had bigger numbers? Wouldn’t we have to continuously find the biggest multiple of seven less than some certain power of ten? No, because they go in a cycle. If seven goes into 9,996 (as we said it does), then it also goes into 99,960. That’s 40 away from 100,000. So we could add 35 (the biggest multiple of seven under 40) to get 99,995 is a multiple of seven. Then seven also goes into 999,950, and hence we can add 49 to get that it goes into 999,999. We’ll be taking one times the millions digit, which means the whole patter will start over again. Three times the ten-millions digit, two times the hundred millions digit, etc.

Let’s work one example with a big number. Is seven a factor of 12,345,678,901,234,567? The pattern for what we must multpliy the digits by, working right to left, is 1-3-2-6-4-5-1-3-2-6-4-5…
That gives
1*7 + 3*6 + 2*5 + 6*4 + 4*3 + 5*2 + 1*1 + 3*0 + 2*9 + 6*8 + 4*7 + 5*6 + 1*5 + 3*4 + 2*3 + 6*2 + 4*1 =
7 + 18 + 10 + 24 + 12 + 10 + 1 + 0 + 18 + 48 + 28 + 30 + 5 + 12 + 6 + 12 + 4 =
245

Is 245 a multiple of seven? I don’t know. Let’s do it again
1*5 + 3*4 + 2*2 = 5+12+4 = 21

21 is a multiple of seven, and so are 245 and 12,345,678,901,234,567.

You can, of course, extend this rule for 11. It goes nicely into 99, 9999, 999999, etc. This gives it the simple rule of take the ones digit plus the ten times the tens digit plus the hundreds digit plus ten times the thousands digit the ten-thousands digit plus ten times the hundred-thousands digit, etc, and that has to be a multiple of 11. You could also think of this as taking every block of two digits and adding them all up. For example 567082 -> 56+70+82 = 288 -> 2 + 88 = 90, so 567082 is not a multiple of 11.

For 13, you could multiply the digits, starting from the right, by 1,10,9,12,3,4,1,10,9,12…

You could figure out similar rules for any other number you pleased, meaning you never actually have to do the division problem to find if one number divides another. That’s not to say applying that algorithm to test whether something’s a multiple of 13 would itself be easy.

Also, about that funny looking rule I spelled out for four and eight. I wrote it in what looked to me like the simplest algorithmic form, but if you practice with a couple you’ll be able to do it without running through that explicit calculation. Note that 2 and 6 are “odd multiples” of two (2*1, 2*3) while 0,4,8 are “even multiples” of 2 (2*0, 2*2, 2*4). For a number to be a mutiple of four, it must pair an even number in the tens digit with an even multiple of two in the ones digit, or an odd number in the tens digit with an odd multiple of two in the ones digit.

Now you can start feeling out even and odd multiples of four. 36 = 4*9 is an “odd multiple”, while 56 = 14*4 is an “even multiple”. After staring at these for a while, I can just kind of feel their even/odd character without actually factoring. If you can do that, the same sort of rule applies for eight as for four. A number is a multiple of eight if its last two digits are an “even multiple” of four and the hundreds digits is even, or if the last two digits are an “odd multiple” of four and the hundreds digit is odd.

Now if you can feel out even and odd multiples of eight, you can use the thousands digit to test if it’s a multiple of 16, and so on to 2^\infty

10 Responses to “Multiples Rule for 7”

  1. Bekka Says:

    Thnx, thiz really helped me w/ my math hw!

  2. Courtney Says:

    can you simplify the multiples of 7 for me it is hard to understand! thank you!

  3. natalie Says:

    where is h 7?????
    D:

  4. Jarvis Says:

    The rule for 7 is really complicated but the rest are really good THANKS!!!

  5. Sam Says:

    Very very good thanks.
    Very much helpful

  6. Divisibility By 7 Revisited « Arcsecond Says:

    […] One of the most-viewed posts I’ve written on this blog is about a rule for checking whether a number is divisible by seven. […]

  7. Mark Eichenlaub Says:

    @Courtney I’ve posted a new version of the rule for seven. https://arcsecond.wordpress.com/2011/02/07/divisibility-by-7-revisited/

  8. camila Says:

    thanks for all but i only need multiples of 7 seven!!!!

  9. camila Says:

    i mean i didn t understand 7

  10. ilovetimon Says:

    thank you :D you’re soooo good at math :X

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