Question

show that 7^n cannot embid with the digit 0 for any natural number n ​

Answer 1

i think the ans is 14

Answer 2

Step-by-step explanation:

Given, 7^n ( where n is any natural no.)

If any no. is going to end with 0 , then they are divisible by 2 and 5

Therefore, in its prime factorization 2 and 5 should occur!

But, 7^n = (7×1)^n

Here, 2 and 5 are not occured.....

Therefore, 7^n never ends with 0

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