Sum

For what value of natural number n, 4^{n} can end with the digit 6?

Advertisement Remove all ads

#### Solution

4^{n} = (2 × 2)^{n} = 2^{n} × 2^{n}

2 is a factor of 4^{n}.

So, 4^{n} is always even and ends with 4 and 6.

When n is an even number say 2, 4, 6, 8 then 4^{n} can end with the digit 6.

**Example:**

4^{2} = 16

4^{3} = 64

4^{4 }= 256

4^{5} = 1,024

4^{6} = 4,096

4^{7} = 16,384

4^{8} = 65,536

4^{9} = 262,144

Concept: Fundamental Theorem of Arithmetic

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads