Question

10^m-10^n is always a factor of 9 if m and n are positive integers?

Can someone please give me a proof for this? I don’t know how to prove that that is true.

Answer 1

See proof in explanation.
The proof works with the modulo operator.

Explanation:

`(10^m-10^n) mod 9`
`= (((10^m) mod 9) - ((10^n) mod 9)) mod 9`
`= ((((10 mod 9)^m) mod 9)-((10 mod 9)^n) mod 9)) mod 9`
`= ((1 mod 9) - (1 mod 9)) mod 9`
`= (1 - 1) mod 9`
`= 0 mod 9`
`= 0`
`=> (10^m - 10^n) " is divisible by 9"`
`=> "9 is a factor of " (10^m - 10^n)`

Answer 2

Please refer to a Proof in the Explanation.

Explanation:

If `m=n,` then `10^m-10^n=0,` which is divisible by `9.`

So, we let, `m ne n, where, m,n in NN.`

Then, either `m > n, or, m < n.`

We will prove the assertion for `m > n; m, n in NN.`

The Proof for `m < n` is similar.

Now, `m,n in NN; m > n rArr m=n+k" for some "k in NN.`

Now, `10^m-10^n=10^(n+k)-10^n=10^n*10^k-10^n, i.e.,`

`10^m-10^n=10^n(10^k-1)=10^n{(9+1)^k-1}.`

Now, expanding `(9+1)^k` using the Binomial Theorem, we have,

`10^m-10^n`

`=10^n[{9^k+""_kC_1*9^(k-1)+""_kC_2*9^(k-2)+...+9+1}-1],`

`=10^n(9^k+""_kC_1*9^(k-1)+""_kC_2*9^(k-2)+...+9),`

`=9(10^n)(9^(k-1)+""_kC_1*9^(k-2)+""_kC_2*9^(k-3)+...+1),`

which clearly shows that `10^n-10^n` is divisible by `9.`

Hence, the Proof.

Enjoy Maths.!