Question

Find the the number 67 n being natural no end with digit 0 digit​

Answer 1

6n=2n x 3n 
For any number ending with 5 the prime factor 5 should be there in its primefactorisation (for eg: 5=5,15=3x5,25=5x5). In all these 5 is a necessary prime number. But 6n has only 2 and 3 as prime factors.So 6n can never end with digit 5 for any natural number n.

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I think you're asking for 6^n. 
No, 6^n cannot end with five.
For - "A ^ n" ( where A is any integer ), to end with 5 ( or have 5 in unit place ).
"A" must be in the form of 2 ^ m X 5 ^ n .
However, here 6 = 2 X 3
ie. 6 cannot be written in the form of 2^m X 5^n.
So, 6 ^n cannot end with five.

And for,  6n.
This too can't end with five, as all multiples of 6 end only with 0, 2,4, 6,8.

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