# Last digit of any power of a number

To find the last digit of any power of a number we should keep following in mind.

1. Last digit of any power number repeats after a certain interval say N. So N is called cyclicity of the number.

2. Last digit of any power of a number = Last digit of any power of the (last digit of original number).

Lets check how does the power of a number repeats itself.

For e.g 2^1 =2, 2^2 =4, 2^3 = 8 , 2^4= 16, 2^5 = 32.

so last digit of 2^1 = 2^5. Hence cyclicity of 2^m = 4. i.e after 4 powers last digit of 2^m will repeat.

Lets find the cyclicity of the integers from 0 to 9.

**0 –>** 0^1 =0. 0^2 =0 *Cyclicity = 1.*

**1 –>** 1^1 =1. 1^2 =1 *Cyclicity =1.*

**2 –>** 2^1 =2, 2^2 =4, 2^3 = 8 , 2^4= 16, 2^5 = 32 Cyclicity =4.

**3 –>** 3^1 =3, 3^2 =9, 3^3 = 27 , 3^4= 81, 3^5 = 243 Cyclicity =4.

**4 –>** 4^1 =4, 4^2 =16, 4^3 = 64 Cyclicity =2.

**5 –>** 5^1 =5, 5^2 =25 Cyclicity =1.

**6 –>** 6^1 =6, 6^2 =36 Cyclicity =1.

**7 –>** 7^1 =7, 7^2 =49, 7^3 = 343 , 7^4= 2401, 7^5 = 16807 Cyclicity =4.

**8 –>** 8^1 =8, 8^2 =64, 8^3 = 512 , 8^4= 4096, 8^5 = 28768 Cyclicity =4.

**9 –>** 9^1 =9, 9^2 =81, 9^3 = 729 , Cyclicity =2.

Lets consider a problem now.

*Q. Find the last digit of 56747^987 ?*

A. So last digit of 56747^987 = Lat digit of 7^987.

Now cyclicity of 7 is 4.

So we divide 987 by 4 and consider the remainder part.

so 987 /4 gives 3 as remainder.

So required last digit = last digit of 7^3 = 3.

In this way we can easily calculate the last digit of any power of a number. 🙂