Question

21. Show that 7" can not end with the digit o for any natural number n.​

Answer 1

AnswEr :

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☯ Prime Factorization :

$:\implies\sf 7^{n} = 1^n \times 7^n$

If any number can end with the digit 0 it must be divisible by 5 & 2.

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Here we can see that the prime Factorisation of the number $\sf\: 7^n$ doesn't contains 5 & 2 as a prime Number.

Now we can say that for any natural number n, the number $\sf\: 7^n$ is not divisible by 5.

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Thus, $\sf\: 7^n$ cannot end with the Digit zero (0) for any Natural number n.

Answer 2

Question ⤵

Show that 7" can not end with the digit o for any natural number n.

Solution ⤵

If 7 is end with digit zero for any natural number 'n'.

then, it was also divisible by 10.

the prime factorisation of 10.

We get => { 10= 2 × 5 }.

But here 7 => 7×1

So We can say that by the Fundamental Theorem of Arithmetic. There is no other factors of 7.

Thus we conclude that 7n is never end with 0