1.

A number in the form of 6^n, where n belongs to natural numbers, can never end with the digit(a) 0(b) 1(c) 2(d) 3The question was asked in an internship interview.My question is taken from Euclid’s Division Lemma in section Real Numbers of Mathematics – Class 10

Answer»

Right option is (a) 0

The best I can explain: If 6^n ends with 0, then it should have 5 as a factor.

In CASE of 6 only 3 and 2 are factors of 6.

Also, from the FUNDAMENTAL theorem of arithmetic, prime factorisation of each number is unique.

Hence, 6^n can never end with 0.



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