Question

$check \: is \: there \: any \: natureal \: number \: n \: for \: which \: {21}^{n?} ends \: with \: zero$

Answer 1

$\sf \star EXPLANATION \\ \\ \implies\star$

$<b>[®]$

If 21ñ ends with o.

then it's must have 7 CommOn Factor.

But we know that the only prime Factor of 21ñ are 3 and 7.

Also.We Know that from fundamental theorem of arithmetic.

that's the prime factorization of each number is unique.

$\So \star$

21ñ can end with 0.

Thanks!!!?