Numbers are quite mysterious. Even the simple notion of natural numbers (1,2,..), used simply for counting has so many emerging properties that have led to the field of Number Theory in Mathematics, with theorems that seem deceivingly simple,which can be written in a single line, understood by a child but take hundreds of pages of proofs or a proof has evaded the greatest mathematicians for centuries, such are the numbers, e.g. Fermat’s last theorem: There exist no solutions for \(x^n+y^n=z^n\), for \(n>2\).
Goldbach’s conjecture: Every integer greater than 2 can be expressed as the sum of two primes.
Collatz’s conjecture: start with any number, if n is even divide by 2, if n is odd multiply by 3 and add 1, this process always stops at 1.
… etc.

I remember my first encounter with being amazed by numbers. It was a simple realization many arrive at, at a young age, i.e. take any integer number greater than 0 multiply by 9, the sum of the digits of the result is 9. This also gives an easy way to check if a number is divisible by 9. For example, 27, 36, 81 etc.
Although this seems magical, the proof is simple:

For a number \(N\) to be divisible by 9 means \(N = 0 \mod 9\). We can write (for 2 digits and similarly for more) \(N = 10n+k\) taking modulus 9, we simplify by subtracting numbers divisible by 9 (i.e. 9, 99, etc. for a number with given number of digits giving \(0 = N \mod 9 = 10n+k \mod 9 = n+k \mod 9\) Therefore the sum of the digits is divisible by 9. If the sum is greater than 9 we repeat the process of summing the digits, and as the largest single digit number divisible by 9 is 9, the sum of the digits will be eventually \(9\). \(\blacksquare\)

Following that, I thought if there are any other simple but non trivial examples like that for other numbers. I stumbled upon \(7\). Take any number take the last digit multiply by 2 and subtract the remaining of the number, if that’s divisible by 7, the number is also divisible by 7. For example, \(28 \rightarrow 2 - 8 \times 2 = 2-16 = -14 = 0\), which is divisible by 7. I’ve arrived at the proof in a similar way, Assuming \(N\) is divisible by 7 we get, \(0=N \mod 7 = 10n+k \mod 7 = 3n +k \mod 7 = 6n+2k \mod 7 = 2k - n \mod 7.\) \(\blacksquare\)