Prove that the difference between numbers with the same sum of digits is a multiple of $9$.

Question

Prove that the difference between numbers with the same sum of digits is a multiple of $9$.

Answer 1

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Well, the difference is not exactly $9$, but it is a multiple of $9$. Consider two $n$-digit numbers $A = a_1a_2\ldots a_n$ and $B = b_1b_2\ldots b_n$ such that $$\sum_{i=1}^n a_i = \sum_{i=1}^n b_i \implies \sum_{i=1}^n (a_i-b_i) = 0$$ You are free to assume the first few digits of the number (or any other for that matter) to be zero, in order to produce numbers that have less than $n$ digits. The proof still goes through.

$$A - B = 10^{n-1}(a_1-b_1) + 10^{n-2}(a_2-b_2) + \ldots + (a_n-b_n)$$ Using $\sum_{i=1}^n (a_i-b_i) = 0$, we can simplify the above expression to get: $$A - B = (10^{n-1}- 1)(a_1-b_1) + (10^{n-2}- 1)(a_2-b_2) + (10^{1}- 1)(a_{n-1}-b_{n-1})$$ Now, $10^k - 1$ is always a multiple of $9$. In addition, each of the terms in the expression of $A-B$ above, is a multiple of $10^k -1$ for some $k \ge 1, k\in\mathbb Z$. As a result, each term of the sum is a multiple of $9$, and therefore $A-B$ itself is also a multiple of $9$.

Why is $10^k-1$ a multiple of $9$ - you may ask? The binomial expansion suffices! $$10^k - 1 = (9+1)^k - 1 = \text{sum of multiples of }9 = \text{multiple of }9$$ I leave the details of the above argument to you. Lastly, this works for all $n\in\mathbb N$, and so we have proved it for all numbers.

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Summary: The difference between two integers with the same sum of digits, is divisible by $9$.