Question

Check whether 6n can end with the digit 0 for any natural number n.​

Answer 1

Let there be any no. n for which 6^n ends with 0.

If 6n ends with 0,

then 5 must be a factor of 6n.

but, 6n = 2n X 3n

So, the uniqueness of fundamental theorem of arithmetic, 5 can't be a factor of 6n.

This contradiction has arisen bcoz of our wrong assumption that 6n ends with 0 .

Therefore, there is no number n for which 6n ends with 0

Answer 2

${\huge{\purple{\underline{\underline{\mathbb{☆Answer☆}}}}}}$

Given :

${6}^{n}$

To prove :

Whether it can end with digit 0 for any natural number n.

Solution :

$∴{6}^{n} = {(2 \times 3)}^{n}$

${\sf{∴prime \: factorisation \: of \: {6}^{n} = {2}^{n} \times {3}^{n} .}}$

${\sf{but, \: for \: {6}^{n} \: to \: end \: with \: digit \: 0, \: prime \: factorisation \: should \: contain \: 2 \: and \: 5 \: in \: it.}}$

${\sf{∴ No \: natural \: number \: n \: for \: which \: {6}^{n} \: ends \: with \: digit \: 0 \: .}}$

Hence, proved ☑️

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# 5 ᴛʜᴀɴᴋꜱ ᴘʟᴢ ❤️