What is the units digit of 367?

The unit digits of 3-base powers follow this sequence: 3, 9, 7, 1
Indeed:
3^1 = 3 => unit digit is 3
3^2 = 9 => unit digit is 9
3^3 = 27 => unit digit is 7
3^4 = 81 => unit digit is 1
3^5 = 243 => unit digit is 3

If 3 is raised to 67th power, the unit digit numbers will repeat this sequence until it stops at 67th power. Because the remainder of the division of 67 by 4 is 3 (16*4 + 3 =67), the digit number of the 67th power will be the third number in the sequence.

So the answer is 7

We get (~3)^1=~3, (~3)^2=~9, (~3)^3=~7, (~3)^4=~1. Units digit repeats 39713. Then, 3^67=3^[4(16)+3] 3^3=~7. Hence, the units digit becomes 7, and the correct answer is D.

Hi Experts,

Can you please explain the below part?

3 (16*4 + 3 =67)

Thanks,

SJ

jain2016 wrote:Hi Experts,

Can you please explain the below part?

3 (16*4 + 3 =67)

Thanks,

SJ

Think of it like this - the pattern for the unit's digit for base 3 goes as follows:

3^1 ---> 3
3^2 ---> 9
3^3 ---> 7
3^4 ---> 1

3^5 ---> 3
3^6 ---> 9
3^7 ---> 7
3^8 ---> 1

So the pattern repeats every four terms. Put another way, every time the exponent is a multiple of 4, the units digit is '1.' Now we can think of a multiple of 4 that's close to 67. (So think either 64, as the above example does, or 68. Let's use 68.)

3^65 ---> 3
3^66 ---> 9
3^67 ---> 7
3^68 ---> 1

Hi David ,

Thanks you sir for your reply. All clear.

Thanks,

SJ

This explanation is very good and easy to follow that i also got D as the answer

The unit digits of 3-base powers follow this sequence: 3, 9, 7, 1
Indeed:
3^1 = 3 => unit digit is 3
3^2 = 9 => unit digit is 9
3^3 = 27 => unit digit is 7
3^4 = 81 => unit digit is 1
3^5 = 243 => unit digit is 3

If 3 is raised to 67th power, the unit digit numbers will repeat this sequence until it stops at 67th power. Because the remainder of the division of 67 by 4 is 3 (16*4 + 3 =67), the digit number of the 67th power will be the third number in the sequence.

So the answer is 7