Prove the following


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


If any number ends with the digit 0 or 5 , it is always divisible by 5 .

If $12^{n}$ ends with the digit zero it must be divisible by $5 .$

This is possible only if prime factorisation of $12^{n}$ contains the prime number $5 .$

Now, $12=2 \times 2 \times 3=2^{2} \times 3$

$\Rightarrow$ $12^{n}=\left(2^{2} \times 3\right)^{n}=2^{2 n} \times 3^{n}[$ since, there is no term contains 5$]$

Hence, there is no value of $n$ e $N$ for which $12^{n}$ ends with digit zero or five.



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