Lizzie came up with a divisibility test for a certain number $m \neq 1$:

Question

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[*] Break a positive integer $n$ into two-digit chunks, starting from the ones place. (For example, the number $354764$ would break into the two-digit chunks $64, 47, 35.$)
[*] Find the alternating sum of these two-digit numbers, by adding the first number, subtracting the second, adding the third, and so on. (In our example, this alternating sum would be $64-47+35=52.$)
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Find $m,$ and show that this is indeed a divisibility test for $m$ (by showing that $n$ is divisible by $m$ if and only if the result of this process is divisible by $m$).

Answer 1

Let's denote each two-digit number as $10a + b,$ where $a$ is the tens digit and $b$ is the units digit. Let $n$ have $k$ two-digit numbers in its decomposition.

When we calculate the alternating sum of these two-digit numbers, we get $10a_1 + b_1 - 10a_2 - b_2 + 10a_3 + b_3 - \cdots \pm 10a_k + b_k,$ where $a_i$ is the tens digit of the $i$-th two-digit number and $b_i$ is the units digit.

This can be rewritten as $10(a_1 - a_2 + a_3 - \cdots \pm a_k) + (b_1 - b_2 + b_3 - \cdots \pm b_k).$ Let $A = a_1 - a_2 + a_3 - \cdots \pm a_k$ and $B = b_1 - b_2 + b_3 - \cdots \pm b_k.$ Then, the sum becomes $10A + B.$

Since $10A + B$ is divisible by $m,$ it implies $10A + B \equiv 0 \pmod{m},$ which gives $10A \equiv -B \pmod{m}.$

We know that $10 \equiv 1 \pmod{m},$ so $10A \equiv A \pmod{m}.$ Hence, we have $A \equiv -B \pmod{m}.$

Therefore, $n$ is divisible by $m$ if and only if $10a + b$ is divisible by $m,$ implying that the divisibility test works for $m = \boxed{11}.$