prove:"an integer is divisible by 9 if the sum of its digits is divisible by 9"?

Question

You are all wrong

Answer 1

m = a(0) + 10a(1) + 100a(2) + ... + (10^n)a(n)
where a(0),...a(n) are the digits of the number m

10 ≡ 1 (mod 9)
10^2 ≡ 1 (mod 9)
...etc.,
10^n ≡ 1 (mod 9) for all natural numbers n

{If you don't know the rules of modulo arithmetic,
note that 10^n-1 is always divisible by 9, since 10^n-1 factored is:
(10-1)(10^(n-1) + 10^(n-2) + ... + 10 + 1)
= 9(10^(n-1) + 10^(n-2) + ... + 10 + 1) }


a(0) + 10a(1) + 100a(2) + ... + (10^n)a(n) ≡ a(0) + a(1)*1 + a(2)*1 + ... + a(n)*1 (mod 9)
≡ a(0) + a(1) + ... + a(n) (mod 9)

so if a(0) + a(1) + ... + a(n) ≡ 0 (mod 9) (i.e., sum of digits is divisible by 9)
Then a(0) + 10a(1) + 100a(2) + ... + (10^n)a(n) ≡ 0 (mod 9)
(i.e., the whole number is divisible by 9)

Q.E.D.

p.s. I'll make a deal with you. If you can prove that my proof is "wrong" then I'll send you $10; otherwise, you send me the 10 points.

Answer 2

well i will give you a helpful hint or 2 :

10^0 mod 9 = 1
10^1 mod 9 = 10 mod 9 = 1
10^2 mod 9 = 100 mod 9 = 1
10^3 mod 9 = 1
...
10^x mod 9 = 1 x an integer >= 0

so a decimal #s representation say for example y= 2376 means we have 2 * 10^3 + 3 * 10^2 + 7 * 10^1 + 6

now take y mod 9 = 2 * 1 + 3 * 1 + 7 * 1 + 6 * 1 mod 9 = 2 + 3 + 7 + 6 mod 9 = 18 mod 9 = 0

now, i leave it to you to formalize the above.

Answer 3

Only thing I would add (to the 1st proof, for example) is that it must be proven that 10^n == 1 mod 9 for all n, even though it seems obvious.

It can be proven by induction. If it's true for 10^m will show it's true for 10^(m+1). It's clearly true for 10^1.

Assume 10^m = 9*j +1 where j is an integer. Then
10^(m+1) =
10*10^m =
10*(9*j+1) =
90*j +10 =
9*(10*j +1) + 1 =
9*k +1 where k = 10*j+1.

QED

Answer 4

Take for example a two digit number.
It can be written as 10a + b. Now it can also be written as:
9a + a + b.
9a is divisible by 9. a+b is divisible by 9. So the number 100a + b is divisible by 9 if a+b is divisible by 9.
Expand this to a number of 3 digits.

Dr. J.

Answer 5

it is jsut the way it is. If the digits of a variety could be further and divided by utilising 3 calmly, it is divisable by utilising 3, so it works for all exponents of three. 9 = 3^2 If the sum of a numbers digits ought to be calmly divided by utilising 27 or 80 one, they could be divisable by utilising 27 or 80 one respectively.

Answer 6

And divisible by 3 as well!

Answer 7

a0 + a1 + a2 +...+ an = 9*k Where k is an integer and a0 a1 a2 etc are the digits of the number.

If we add 9a1 + 99a2 + 999a3 +...+(10^n - 1)an to the above we get the number itself.

Notice that adding 9a1+99a2+999a3 + ... + (10^n-1) to 9*k results in an integer that is divisible by 9

Therefore the number is also divisible by 9

Answer 8

162 is divisable by 9 because 1+6+2=9 and 162/9=18