# How to prove divisibility of 11 by basic algebra? (Without modal arithmetic or proof by induction) [duplicate]

Test is defined as: N is divisible by 11 iff the difference between the two sums of the odd and even-numbered digits is divisible by 11.

So I actually need 2 proofs for (1) if alternating sum is divisible by 11, then N is divisible by 11; and (2) if N is divisible by 11, then alternating sum is divisible by 11.

My teacher did the following and although it works, it also seems unnecessary at times and extremely long.

My questions are:

(1) is there a less convoluted way to prove the above?

(2) similarly, how do I prove N is divisible by 11 when the alternating sum of its digits is divisible by 11?

LEMMA 1: $10^{2n}-1$ is a multiple of $11$ for $n=0,1,2,\dots$

Proof: Since $\dfrac{100^n - 1}{100-1} = \dfrac{10^{2n}-1}{99}$ is an integer, it follows that $11$ divides $10^{2n}-1$.

LEMMA 2: $10^{2n+1}+1$ is a multiple of $11$ for $n=0,1,2,\dots$

Proof: $(10^{2n}-1) + (10^{2n+1}+1)=10^{2n} + 10^{2n+1} = 10^{2n}(1+10) = 10^{2n}\cdot 11$. Since $10^{2n}\cdot 11$ and $(10^{2n}-1)$ are multiples of $11$, then so too is $(10^{2n+1}+1)$.

I will state the final theorem in terms of $4$-digit numbers. The generalization to numbers of any size should be obvious.

THEOREM: Let $N = a_310^3 + a_210^2+a_110+a_0$ be any $4$-digit number. Then $N$ is a multiple of $11$ if and only if $(a_2+a_0) - (a_3 + a_1)$ is a multiple of $11$.

Proof: \begin{align} N &= a_310^3 + a_210^2+a_110+a_0 \\ &= [a_3(10^3+1) - a_3] +[a_2(10^2-1) + a_2] +[a_1(10+1) - a_1] + a_0 \\ &= [a_3(10^3+1) + a+2(10^2-1) + a_1(10+1)] + (a_2+a_0)-(a_3+a_1) \end{align}

Since $a_3(10^3+1) + a+2(10^2-1) + a_1(10+1)$ is a multiple of $11$, then $N$ is a multiple of $11$ if and only if $(a_2+a_0)-(a_3+a_1)$ is a multiple of $11$.