Question

show that any number of the form 6×,€N can never end with the digit 0​

Answer 1

The number is of the form $6^{n}$, where $n\in\mathbb{N}$.

Now, $6^{n}$

$\quad=(3\times 2)^{n}$

$\quad=3^{n}\times 2^{n}$

Here n > 0 and we also know that no exponential value of 2 or 3 ends with a 0.

Reason:

$\quad\quad 2^{n}=2,4,8,16,32,...$

$\quad\quad 3^{n}=3,9,27,81,243,...$

Therefore, $6^{n},\:n\in\mathbb{N}$ is not divisible by 10, and a number which is not divisible by 10, cannot end with a 0.

This completes the proof.

Read more on Brainly:

1. The number 3 13 -3 10 is divisible by - https://brainly.in/question/379428
2. Prove that 10 n + 3.4 n+2 + 5 is divisible by 9 by Principle of Mathematical induction. Plz Plz Plz answer fast. - https://brainly.in/question/1062234

Answer 2

If the number 6^x ends with the digit zero, then it should be divisible by 5.

Prime factorisation of 6^x = (2×3)^x

Since the prime factorisation of 6^ x doesn't contain prime number 5, therefore it clearly states that 6^x, x € N is not divisible by 5 and hence cannot end with a zero.

PROVED ABOVE

THANK YOU

HOPE IT HELPS...