1. ## Dividing Integers

I have run across some interesting properties of the quotient obtained from dividing 2 integers. I will list the properties in which I am interested at the conclusion of this post. The question is: Where can I find proofs of these properties.

Here are the properties:

1. An integer is divisible by 3 if the sum of its digits is divisible by 3.

2. An integer is divisible by 4 if the last 2 digits of the number form a number which is divisible by 4.

3. An integer is divisible by 8 if the last 3 digits of the number form a number which is divisible by 8.

4. An integer is divisible by 9 if the sum of its digits is divisible by 9.

Thanks for any input.

... doug

2. ## Re: Dividing Integers

The proof of 1), for a three-digit number "abc"
abc = 100a + 10b + 1c
= 99a + 9b + (a + b + c)
3 divides 99a + 9b, so 3 divides "abc" iff 3 divides (a + b + c), the sum of the digits.
You can do this for numbers of any length, but I'm not latexing that from my iPhone.

2). Express any number as "abcdef00" + "gh" (for example).
Since "abcdef00" is a multiple of 100, it is divisible by 4. So all that remains is to check "gh", the last two digits.
3) same as 2), since 8 divides 1000
4) same as 1

3. ## Re: Dividing Integers

Awesome! Thanks.

... doug

4. ## Re: Dividing Integers

You can work out other divisibility tests for, say, prime numbers (the building blocks on the integers, if you will).
It gets complicated quickly, though!
Consider divisibility by 7. We look for multiples of 7 that are CLOSE TO POWERS OF TEN (ten, since we use base ten).
It gets nasty quickly, but stick with me!
1 + 0 = 1
3 + 7 = 10
2 + 98 = 100
(-1) + 1001 = 1000
(-3) + 10003 = 10000
(-2) + 100002 = 100000
The second number in each equation is divisible by 7. So for a six digit number "abcdef", we write this as:
a*100000 + b*10000 + c*1000 + d*100 + e*10 + f*1.
=
(a*100002 + b*10003 + c*1001 + d*98 + e*7) + [-2*a - 3*b - 1*c + 2*d + 3*e + 1*f]
Now everything in (...) is divisible by 7. It remains to check that the contents of [...] are divisible by 7. It's not as clean as just adding up the digits, but you have a method.
p.s. the pattern of coefficients in the brackets [-2,-3,-1,2,3,1] repeats so you can extend this to arbitrarily large numbers.

5. ## Re: Dividing Integers

Now, see if you can work out something similar for divisibility by 11. It's fairly simple, since
1 + 99 = 100
(-1) + 1001 = 1000
1 + 9999 = 10000
(-1) + 100001 = 100000, and so on.

6. ## Re: Dividing Integers

To summarize (2) (3): 2^n divides a number if the last n digits can be divided by 2^n.

7. ## Re: Dividing Integers

Where can I find proofs of these properties?
/summary

8. ## Re: Dividing Integers

For number 4: I don't have time to prove it at the moment, but you can recursively add the digits of a number divisible by 9 to get the number 9. That might be a hint for the proof. I'll work on it too.