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Lizzie came up with a divisibility test for a certain number :

Break a positive integer into two-digit chunks, starting from the ones place. (For example, the number would break into the two-digit chunks )

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 )

Find and show that this is indeed a divisibility test for (by showing that is divisible by if and only if the result of this process is divisible by ).

can you please make the explanation simple and easy to understand

Dysoncloud May 17, 2024

#2**0 **

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 $m = \boxed{11}.$

LiIIiam0216 May 18, 2024