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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.

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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?

user64742
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AgeOfTheGeek
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1 Answers1

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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 $11$.

Steven Alexis Gregory
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