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# SSC CGL Questions on Remainder PDF

Download SSC CGL  Remainder Questions with answers  PDF based on previous papers very useful for SSC CGL exams. Very Important Remainder Questions for SSC exams

Question 1: The difference between two numbers is 1146. When we divide the larger number by smaller we get 4 as quotient and 6 as remainder. Find the larger number.

a) 1526

b) 1431

c) 1485

d) 1234

Question 2: A number between 1000 and 2000 which when divided by 30, 36 & 80 gives a remainder 11 in each case is

a) 1451

b) 1641

c) 1712

d) 1523

Question 3: What is the highest number which when divides the numbers 1026, 2052 and 4102, leave remainders 2, 4 and 6 respectively.

a) 512

b) 1024

c) 128

d) 256

Question 4: Let x be the smallest number greater than 600 which gives the remainders 2, 3 and 4, when divided by 5, 6 and 7, respectively. The sum of digits of x is:

a) 14

b) 15

c) 13

d) 16

Question 5: When 5, 6, 8, 9 and 12 divide the least numberx, the remaindereach caseis 1, but x is divisible by 13. What will be the remainder when x is divided by 31?

a) 0

b) 1

c) 3

d) 5

Question 6: Let x be the greatest number which when divides 6475, 4984 and 4132, the remainder in each case is the same. What is the sum of digits of x ?

a) 4

b) 7

c) 5

d) 6

Question 7: A number, when divided by 114, leaves remainder 21. If the same number is divided by 19, then the remainder will be

a) 1

b) 2

c) 7

d) 17

Question 8: A number, when divided by 136, leaves remainder 36. If the same number is divided by 17, the remainder will be

a) 9

b) 7

c) 3

d) 2

Question 9: When ‘n’ is divided by 5 the remainder is 2. What is the remainder when n2 is divided by 5?

a) 2

b) 3

c) 1

d) 4

Question 10: If $17^{200}$ is divided by 18, the remainder is

a) 1

b) 2

c) 16

d) 17

Question 11: The least multiple of 13 which when divided by 4, 5, 6, 7 leaves remainder 3 in each case is

a) 3780

b) 3783

c) 2520

d) 2522

Question 12: A number x when divided by 289 leaves 18 as the remainder. The same number when divided by 17 leaves y as a remainder. The value of y is

a) 2

b) 3

c) 1

d) 5

Question 13: The least number which when divided by 6, 9, 12, 15 and 18 leaves the same remainder 2 in each case is :

a) 180

b) 182

c) 178

d) 176

Question 14: For any integral value of n, $3^{2n}$ + 9n + 5 when divided by 3 will leave the remainder.

a) 1

b) 2

c) 0

d) 5

Question 15: Find the least number which when divided by 12, 18, 36 and 45 leaves the remainder 8, 14, 32 and 41 respectively.

a) 186

b) 176

c) 180

d) 178

Let the smaller number be $x$ and the larger number = $(x+1146)$

According to ques, on dividing the larger term by smaller one,

=> $(x+1146)=4x+6$

=> $4x-x=1146-6$

=> $3x=1140$

=> $x=\frac{1140}{3}=380$

$\therefore$ Larger number = $380+1146=1526$

=> Ans – (A)

LCM of given 3 numbers (30, 36, 80) = 720

Multiple of 720 between 1000 and 2000 is 1440.

$\therefore$ Number which gives a remainder 11 in each case (1440 + 11) = 1451

Hence, option A is the correct answer.

Let the given number be x
Let a be the quotient when x is divided by 114
So $\frac{x}{114}$ = a$\frac{21}{114}$
so x = 114a + 21
when x is divided by 19 it can be written as
$\frac{x}{19} = \frac{114a + 21}{19}$
114 is divisible by 19 and 21 leaves a remainder of 2.

Number will be (136n + 36) where n is quotient
hence when it is divided by 17 remainder for $\frac{136n +36}{17}$ will be 2 as 136 is divisible by 17 and 36=34+2

n = 5k+2 (where k is quotient )
so $n^2 = 25k^2 + 4 + 20k$
Now when $n^2$ will divided by 5 , remainder will be 4.

$17^{200}=(18-1)^{200}$
Hence, when it is divided by 18, the reminder equals $(-1)^{200}=1$

Number will be equal to 420t +3 = 13M
put values of M and t accordingly and find least value of it.

The number is of the form 289n+18.
Which is equal to 17*(17n+1) +1
So, when the number is divided by 17, the reminder is 1

The numbers 6,9,12,15,18 leaves same remainder 2 in each case.

So, what we need to do is find the L.C.M. of these numbers and add 2 to it

=> L.C.M. of 6,9,12,15,18 = 180

=> Required no. = 180+2 = 182

Expression = $3^{2n}$ + 9n + 5

= $3^{2n}$ + 9n + 3 + 2

Taking 3 common from each term, we get :

=> 3 ($3^{2n-1}$ + 3n + 1) + 2

Now, if we divide the above term by 3, remainder will be 2.