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Published 1997 Revised 2021
Here is another method for testing divisibility by $7$ suggested by T.R.Mukundan
In step 1 of the method, we used the fact that there is a unique number in the $7$ times table (up to $10 \times 7 = 70$) ending in each digit $0$ to $9$. Is this true in other times tables? It works for $3, 7,9, 11\ldots$. But it certainly doesn't work for $2$ - no multiple of $2$ ends in $1$. It turns out that it works for any number not divisible by $2$ or $5$ (the prime factors of $10$). To understand why this is true, read this article on the Chinese Remainder Theorem.
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
How many noughts are at the end of these giant numbers?
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.