Divisibility Problem

Is 59136 divisible by 2?

Of course it is. And you know this because the last digit is a 6.

Is 558162 divisible by 11? Not so easy. If you your schooling was like mine, you never learned a quick trick to check for divisibility for 11. Check this out:

558162

5+8+6 = 19

5+1+2 = 8

198 = 11

Since 11 is divisible by 11, 558162 is divisible by 11. Go ahead, check your calculator!

Here is a challenge problem using the new for 11. What digit could you insert in the number 43160523 to make it divisible by 11? And where would you insert it?

First, we should ensure that the number is not already divisible by 11.

43160523

4+1+0+2 = 7

3+6+5+3 = 17

717 = -10

-10 is not divisible by 11. The nearest multiples of 11 are -11 and 0. We need to insert a digit somewhere to modify the number to get a total of -11 or 0. To get to -11, we could increase the sum of 17 to 18. One way to do this would be to insert a 1 at the beginning of the number. Hence, 143160523 would be our new number. You can verify that it is divisible by 11. If you came up with a different solution, please post it in the comments!

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2 Comments

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2 responses to “Divisibility Problem

  1. Matias Gervai

    That’s cool, I didn’t know the rule for divisibility by 11.
    Another related trick involves divisibility by 3; you may or may not know about this already…
    I’ll start with something you all probably know.
    In order to tell whether a number is divisible by 3, as most of us are taught, we add the digits in the number, and see if the sum of its digits is divisible by 3. If it’s too large, we can continue to do so again and again until it’s small enough to tell whether it’s divisible by 3.
    For example, is 23948759 divisible by 3?
    Well, 2+3+9+4+8+7+6+9 = 48
    4+8 = 12
    1+2 = 3
    3 is divisible by 3! (and so were 48, and 12)

    What fewer people know however, is that this number is called the “digital root”, and you can do some other cool stuff with it.
    First, one can get to the digital root of a number much quicker by using the elimination of 9’s method. That is, one can just chuck the 9’s out, so our above example would become

    2+3+4+8+7+6 = 30
    3+0 = 3!

    And extending that idea, if you can make “9”s with the numbers, you can get there even faster. For example, if we steel one from the 7, and add it to the 8, and combine the 3 and the 6, we can quickly make 2 9’s and get rid of them. So then
    2+3+4+8+7+6 –> 2+4+(8+1)+(7-1)+(6+3) Chucking the 9’s yields
    2+4+6 = 12 –> 1+2 = 3!

    With very little practice, one can become VERY quick at finding the digital root.
    Why would you want to know the digital root?
    Because you can use it to quickly check your math!

    Say I want to add 2 big numbers by hand. Like 2394876 + 1234087
    I go through it, and get 3628963. Did I do it right?
    Well, do the same operation with the digital roots, and it should all match up:
    2394876 –> 3,6,9 go away, leaving 2+4+(7-1)+(8+1), –> Digital root = 3
    +1234087 –> 8,1,7,2 go away leaving 3+4 = 7
    =3628963 –> 9,3,6,3,6 go away leaving 2+8=10

    7+3=10, so I did my math right!

    Of course, it could be that you do it wrong, and the digital roots still happen to add to the same, but that’s quite unlikely, so it’s a quick good level of safety.
    It also works with subtraction, multiplication and (albeit with a little more work), division.
    I leave exploring the other operations to you!

    • Mike Licina

      I noticed that in the original post one of the additions is incorrect. 5+1+2=8, not 7. This would make the subtraction 19-8=11, which is correct, so no harm, no foul. Just thought you’d like to know.

      Also worthy to note is the divisibility pattern for powers of 2. To know whether a number is divisible by 2^n, just check the final n digits for divisibility. For example, the number 8722256 is divisible by 8 (2^3) since 256=8*32. The inductive proof follows:

      Let p=a*10+b. By the rules of divisibility, we know that if d|a or d|b, then d|ab. Since 2|10, it follows that 2|a*10, and therefore 2|p-b. We then consider two possible cases. If 2|b, then 2|p, since if d|m and d|n (when m>=n), then d|(m-n). If b is indivisible by 2, then p is indivisible by 2 by the same property. Thus, if 2|b, then 2|p. This proves the case of n=1.

      Now, consider the n+1 case. We want to prove that if 2^(n+1) divides the last (n+1) digits, then 2^(n+1)|p. Rewrite p as A*10^(n+1)+B. Since 2|10, it follows that 2^(n+1)|10^(n+1), and therefore 2^(n+1)|a*10^(n+1)=p-B. Therefore, if 2^(n+1)|B, then 2^(n+1)|p. If, however, 2^(n+1) does not divide B, then 2^(n+1) does not divide p.

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