given,
6^n
if 6^n is to be end with for a natural number n it should be divisible by 2 & 5
prime factorisation of 6^n should contain the prime numbers 2 & 5 but 6^n contains only both in its prime factorisation...
[6^n=(2×3)^n)
since 5 is not present in the prime factorisation there is no natural number "n"for which 6^n ends with digit 0.
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Answers & Comments
given,
6^n
if 6^n is to be end with for a natural number n it should be divisible by 2 & 5
prime factorisation of 6^n should contain the prime numbers 2 & 5 but 6^n contains only both in its prime factorisation...
[6^n=(2×3)^n)
since 5 is not present in the prime factorisation there is no natural number "n"for which 6^n ends with digit 0.