Question

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) 3

The 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 1

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.