Why does every number of shape ababab is divisible by $13$?

Note that $$ [ababab] = a\times 101010 + b \times 10101 = 13 (7770a + 777b) $$

Also noteworthy: $$ 10101 = 1 + 10^2 + 10^4 \equiv \\ 1 + 3^2 + 3^4 = 1 + 9 + 9^2 \equiv\\ 1 +(-4) + (-4)^2 = 1 - 4 + 16 = 13 \equiv 0 $$ where $\equiv$ indicates equivalence modulo $13$.


These numbers are of the form $(10a+b)\cdot 10101$, and $10101=13\cdot 777$.


Note: $$ab=a\times 10+b\times 1$$

so $$ab\times 100=ab00$$ therefore

$$abab=ab00+ab=ab(100+1)=ab\times 101$$ $$abab00=ab\times101\times100=ab\times10100$$ $$ababab=ab\times10101=ab\times3\times7\times13\times37$$

Hence it is divisible by $3$,$7,13$ and $37$.

Tags:

Arithmetic

Elementary Number Theory

Divisibility

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