Question

number 6n can not end with(where n is a natural number)
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Answer 1

Answer:

Let us on the contrary assume that 6^n can end with a digit 0.

But, 6 = 2¹×3¹

It cannot be expressed in the form 2^x×5^y,

where x,y>0.

Therefore, 6^n cannot end with a digit 0.

THAT'S YOUR ANSWER↑

BTW, ARE YOU IN 10th?