Question

Question 11
Show that ${12}^n$ cannot end with the digit 0 or 5 for any natural number n.

Answer 1

If any number ends with the digit 0 or 5, it is always divisible by 5.
If ${12}^n$ ends with the digit zero or five, it must be divisible by 5.
This is possible only if prime factorization of 12 contains the prime number 5.
Now, $12=2\times 2\times 3=2^2\times 3$
$\Rightarrow$ ${12}^n=(2\times 2\times 3{)}^n=2^n\times 2^n\times 3^n$
${12}^n$ is not divisible by 5 as 5 is not in the prime factorisation of ${12}^n$.
Hence, there is no value of $nϵN$ for which ${12}^n$ ends with digit zero or five.