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.

2 Answers
Dec 6, 2017

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)#

Dec 6, 2017

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.!