Question

show that 12^n cannot end with the digit 0 or 5 for any natural number n​

Answer 1

we know that any positive integer ending with 0 is divisible by 5 and so it's prime factorisation must contain the prime no.5

here 12^n={3×2×2}^n

3 and 2 are the prime no . 8n the prime factorisation of 12^n by uniqueness of the fundamental theorm of arithmetic 5 do not occur in the prime factorisation of 12^n for any n9.{n}

:12^n does not end with 0

may this help u shravni

Answer 2

$\huge\bf\purple{\mid{\overline{\underline{Answer}}}\mid}$

Here, Number=12n where n stand for any natural number .

Now 12n= (2n x3)n

Now , For 12n to end with 0, it should have 2 as well as 5 in its Prime factors to end with 0, Also to end with 5 , it requires at least a single multiple of 5 in its Prime Factors, So 12n cannot end with the digit 0 or 5.

${\bold{\red{\huge{Hope\: it \:helps}}}}$