The prime blocks of 6, 6² and 6³ are shown, and below them the blocks of 10, a number ending in 0. Can 6ⁿ end with the digit 0 for any natural number n?

Question

Accepted Answer

No: 6ⁿ has only the primes 2 and 3, but a number ending in 0 needs the prime 5. 6^n = 2^n × 3^n. A number ending in 0 is divisible by 10 = 2 × 5, so 5 would have to be one of its primes. By uniqueness of prime factorisation, 5 never appears, so 6^n never ends in 0.

The prime blocks of 6, $6^{2}$ and $6^{3}$ are shown, and below them the blocks of 10, a number ending in 0. Can 6ⁿ end with the digit 0 for any natural number n?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Fundamental Theorem of Arithmetic practice

The prime blocks of 6, 62 and 63 are shown, and below them the blocks of 10, a number ending in 0. Can 6ⁿ end with the digit 0 for any natural number n?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Fundamental Theorem of Arithmetic practice

The prime blocks of 6, $6^{2}$ and $6^{3}$ are shown, and below them the blocks of 10, a number ending in 0. Can 6ⁿ end with the digit 0 for any natural number n?