# Easy Number System Solved QuestionAptitude Discussion

 Q. Find the remainder when $2^{31}$ is divided by $5$.
 ✖ A. 4 ✖ B. 5 ✔ C. 3 ✖ D. 7

Solution:
Option(C) is correct

$2^{10} = 1024$

Unit digit of $2^{10} × 2^{10} × 2^{10}$ is 4 [as $4 × 4 × 4$ gives unit digit 4].

⇒ Unit digit of 231 is 8. Now, 8 when divided by 5, gives 3 as remainder.

Hence, 231 when divided by 5, gives 3 as remainder.

Edit: For an alternative solution, check comment by Kumar Vikram.

Edit 2: For yet another alternative solution involving 'modulus' function, check comment by Sravan Reddy.

## (5) Comment(s)

AKARSHI
()

we can also do like this:-

2^31 = (2^2)^15 * 2^1

= (-1)^15 * 2

= -1*2 = -2

2^31/5 = -2/5 = 3

Krishna
()

can u solve remainder when 128^1000 devided by 153? please by modulus method

Sravan Reddy
()

I think learning '$\text{mod}$' or '$\text{modulus}$' function would be helpful to solve all reminder related problems. Especially very complex problems involving big numbers.

$2 = 2\text{mod}5$

$2^4=16\text{mod}5 = 1\text{mod}5$

$(2^4)^7=(1^7)\text{mod}5$

Multipying with $2^3$ on both sides,

$=2^{(28)}*2^3=2^3*1\text{mod}5$

$2^{31}=8\text{mod}5 = 3\text{mod}5$

Hence, the reminder is $3$.

It might look greek and latin at first but it would be your favourite tool for all reminder related problems once you learn it.

All the above steps could be done in mind in 20-30 secs if you master the 'mod' tool.

Yashwant
()

Could u do this whit ((2^3)^10)×2^1

Then remainder is 1 why ?

Kumar Vikram
()

We can also do like this.

$2^{31}$ can be written as $(2^4)^7 * 2^3$

Now when $2^4$ is divided by 5, the remainder is 1 and when $2^3$ is divided by 5 the remainder is 3.

Therefore answer $= 1*3 =\textbf{3}$.