Tests of divisibility - 3

A small variation in the tests described in the last two posts yields more findings.


Instead of the mapping $10a + b \mapsto a - k b$, what if we have $10a + b \mapsto a + k b$ (that is, with a plus sign replacing the minus sign)? 


Nearly the same steps lead us to see that this yields a test for divisibility by $10k - 1$.


Thus, we consider the following function $g_k$ which acts on the positive integers $x$ as follows: If $x = 10a + b$, where $0 \le b \le 9$, then $g_k (x) = a + k b$. Then we claim that $x$ is divisible by $10k - 1$ if and only if $g_k (x)$ is divisible by $10k - 1$.


To see that this claim is valid, observe that
\[k x - g_k (x) = k(10a + b) - (a + kb) = a(10k - 1).\]
Thus, $k x - g_k (x)$ is a multiple of $10k - 1$. Now:

• If $x$ is divisible by $10k - 1$, then so is $k x$, and so also is $g_k (x)$ from the above relation.
• If $g_k (x)$ is divisible by $10k - 1$, then so also is $k x$, from the above relation. And since $k$ and $10k - 1$ are coprime, this means that $x$ itself is divisible by $10k - 1$.

It follows that $x$ is divisible by $10k - 1$ if and only if $g_k (x)$ is divisible by $10k - 1$. This proves our claim.


Just like earlier we get several tests of divisibility from this fact, all at the same time.

• Putting $k = 2$, we get a test of divisibility by $19$: The integer $10a + b$ is divisible by $19$ if and only if $a + 2b$ is divisible by $19$. So the test is: Add twice the units digit to the rest of the number, and proceed as earlier.
• Putting $k = 3$, we get a test of divisibility by $29$: The integer $10a + b$ is divisible by $29$ if and only if $a + 3b$ is divisible by $29$. So the test is: Add three times the units digit to the rest of the number, and proceed as earlier.
• With $k = 4$ we get a test for divisibility by $13$ (which is a divisor of $39$), namely: The integer $10a + b$ is divisible by $13$ if and only if $a + 4b$ is divisible by $13$. So the test is: Add four times the units digit to the rest of the number, and proceed as earlier.
• Continuing, $k = 5$ yields another test for divisibility by $7$ (because $49 = 7 \times 7$), and $k = 6$ yields a test for divisibility by $59$. 
• Of particular interest is the case $k = 10$, for then $10k - 1 = 99$, which factorizes as $9 \times 11$. So this is simultaneously a test for divisibility by $9$ as well as by $11$. The test is: Add ten times the units digit to the rest of the number, and proceed as earlier. Then the original number is divisible by $9$ if and only if the final number is divisible by $9$, and the original number is divisible by $11$ if and only if the final number is divisible by $11$.

Quite a rich set of findings, all of which emerge from such a simple observation!