Mar 26, 2013 · The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three.

Jul 23, 2012 · Recognize that divisibility by a 17 means you can write the number as 17n for a positive integer n. So our statement is if x-5y=17n then 10x + y = 17m , where x,y,m,n are positive integers. Assume x - 5y = 17n .

Sep 27, 2021 · Proof of a divisibility rule. 5. divisibility for numbers like 13,17 and 19 - Compartmentalization method. 1.

Yet another way to do it is to use a similar alternating sum test as for divisibility by 11, but in 3 digit groups, subtracting first, with the sum's divisibility by 7 determining the original number's divisibility by 7. In the $3201828$ example, this is $828 - 201 + 3 = 630,$ so since $630$ is divisible by $7$, so is $3201828$. I got this from ...

The rule $7 \mid 10n+d \iff 7 \mid n-2d$ works fine when the number $10n+d$ is "small" but it would be too much work for, say, the number $2726394118$ However, the fact that $7 \mid 1001$ can be used to speed things up a bit.

This results in another rule: 10^x/3 = +1 which is the rule for divisibility by 3. Now consider the RULE FOR 7 A^x/B= C. Here A=10 (base 10), B = 7. Now we have to choose values for x such that it gives a remainder of C= +1 or -1. we see the following pattern: 10^0/7 = +1. 10^3/7 = -1. 10^6/7 = +1. So we see that +1 and -1 remainders alternate ...

I tried using $10\equiv 3\pmod 7$ like the other divisibility rule proofs, but I don't think it works the same way for 7. A hint given in the book says 1001=(7)(11)(13) elementary-number-theory

Everyone knows the divisibility rule of $13$. Test for divisibility by $13$: Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. For Example : $$50661$$ $$5066+4=5070$$ $$507+0=507$$ $$50+28=78$$

Feb 23, 2020 · Divisibility by 73 and by 137 is tested with alternating sums of four-digit groups. The number 137 is the largest prime that can be tested using simple sums of alternating sums involving four or fewer digits.

This seems to be a poor man's extension to the decimal divisibility by $9$ rule. The answer to the above mentioned question, says that "One needn't memorize motley exotic divisibility tests. Motley exotic divisibility tests are very welcome here!