We need to find What is the remainder when \(3^{283}\) is divided by 5

Theory: Remainder of a number by 5 is same as the unit's digit of the number

(Watch this Video to Learn How to find Remainders of Numbers by 5)

Using Above theory , Let's find the unit's digit of \(3^{283}\) first.

We can do this by finding the pattern / cycle of unit's digit of power of 3 and then generalizing it.

Unit's digit of \(3^1\) = 3
Unit's digit of \(3^2\) = 9
Unit's digit of \(3^3\) = 7
Unit's digit of \(3^4\) = 1
Unit's digit of \(3^5\) = 3

So, unit's digit of power of 3 repeats after every \(4^{th}\) number.
=> We need to divided 283 by 4 and check what is the remainder
=> 283 divided by 4 gives 3 remainder

=> \(3^{283}\) will have the same unit's digit as \(3^3\) = 7
=> Unit's digits of \(3^{283}\) = 7

But remainder of \(3^{283}\) by 5 cannot be more than 5
=> Remainder = Remainder of 7 by 5 = 2

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Remainders

Using the multiplication law for remainders


3^4(70) x 3^3/5

(3^4)70/5 x 3^3/5

1 x 2 = 2

Answer C

Adewale Fasipe, GRE, GMAT instructor from Lagos, Nigeria.

3 follows the pattern as 3 , 9, 27, 81 , 243 so after multiplication 4 times the pattern of units place digit repeats so 283 divided 4 will have remainder 3 means the number will have units place digit as 7 because as we follow the pattern of 3, 9, 27, 81 ... the 3rd place number's units digit is 7. Now dividing this number by five will have remainder 2.