Is the number 6rast power n ends with digit 0 for any natural 'n' ? justify your answer
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If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6^n = (2 ×3)^n
It can be observed that 5 is not in the prime factorisation of 6^n.
Hence, for any value of n, 6^n will not be divisible by 5.
Therefore, 6^n cannot end with the digit 0 for any natural number n.
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If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6^n = (2 ×3)^n
It can be observed that 5 is not in the prime factorisation of 6^n.
Hence, for any value of n, 6^n will not be divisible by 5.
Therefore, 6^n cannot end with the digit 0 for any natural number n.
✌✌✌✌✌
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