CAT Remainder Questions PDF [Most Important with solutions]

CAT Questions on Remainder have appeared in the CAT Quantitative Ability section. You can check out the questions on Remainders from CAT Previous year papers. In this article, we will look into some important CAT Questions on Remainders and remainder theorem. These are a good source for practice; If you want to practice these questions, you can download these CAT Questions on Remainder PDF, which is completely Free.

Remainder CAT Questions are a part of CAT Quant Number Systems. Applications of Remainders questions in CAT were asked in the past years. Take a Free CAT mock test and also try to solve CAT previous questions to get good understanding of these CAT Questions.

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Question 1: Let $f(n)=n^{log_{24}37}$. Find the remainder when $f(2)\times f(3) \times f(4)^{2}\times f(6)$ is divided by 13.

a) 4

b) 7

c) 1

d) None of these

1) Answer (A)

Solution:

$f(2)\times f(3) \times f(4)^{2}\times f(6)$
= $2^{log_{24}37}\times 3^{log_{24}37}\times 16^{log_{24}37}\times 6^{log_{24}37} $
=$576^{log_{24}37}=24^{2{log_{24}37}}=24^{log_{24}37^{2}} = 37^{2} $
We have to find the remainder when $37^{2} $ is divided by 13.
$37^{2} =(39-2)^{2}$
Thus, the remainder will be $(-2)^{2}=4$

Question 2: What is the remainder when $120!$ is divided by $120^{24}$ ?

2) Answer: 0

Solution:

The product of any five consecutive numbers is divisible by 5!
So, 1*2*3*4*5 is divisible by 120.
Similarly, 6*7*8*9*10 is also divisible by 120.
120! can be written as (1*2*3*4*5)*(6*7*8*9*10)*…*(116*117*118*119*120)
Each of the 24 terms is divisible by 120
Hence, the product is divisible by $120^{24}$
So, the remainder when $120!$ is divided by $120^{24}$ is 0.

Question 3: What is the remainder when 3232323232……300 numbers is divided by 37?

3) Answer: 0

Solution:

We know that 37 X 3=111
111 has a unique property. The remainder of a number when divided by 111 is same as the remainder obtained when the sum of the numbers when arranged 3 at a time is divided by 111.
So, the sum will be- [323]+[232]+[323]+[232]…….100 times or [555]+[555]+…50 times which will be 555 X 50 which is divisible by 111. Thus, 37 which is a factor of 111 will also divide the number. Thus, the answer is 0.

Question 4: A function $F_n$ is defined as $F_n = 11^n + 13^n$. What is the remainder when $F_{105}$ is divided by 144.

a) 72

b) 108

c) 96

d) 120

4) Answer (A)

Solution:

We need to calculate the remainder when $11^{105} + 13^{105}$ is divided by 144.
Note that this equals $(12-1)^{105} + (12+1)^{105}$ being divided by 144.
$(12-1)^{105}=12^{105} – 105\times12^{104}+…+105\times 12 -1$
$(12+1)^{105}=12^{105} + 105\times12^{104}+…+105\times 12 +1$
The even terms of both the expressions cancel each other.
Hence, the only term where the power of $12$ is less than 2 is $105 \times 12 + 105\times 12=210 \times 12 = 2520$
The remainder when this is divided by 144 is 72

Question 5: P is the smallest multiple of 45 such that every digit of P is either 7 or 0. What is the remainder when P is divided by 11?

a) 4

b) 10

c) 0

d) 6

5) Answer (A)

Solution:

P is a multiple of 5 and 9.
As every digit of P is either 7 or 0, we can conclude that the last digit of P is 0
Similarly, the sum of the digits of P is divisible by 9.
Hence, we can conclude that P = 7777777770 (nine 7s followed by a 0)
As 77 is a multiple of 11, the remainder when 7777777770 is divided by 11 is equal to the remainder when 70 is divided by 11 and equals 4

Question 6: A group of 20 friends went to a party. On their way they found a certain number of gold coins. The number of coins is such that if they divide them equally, then 18 coins will remain. Had there been 1 friend less, 17 coins would have been left after dividing all the coins equally. Similarly, if the number of friends had been 18, the number of coins left after equal division would have been 16 and so on. Also, it is known that the number of coins is the least possible value that satisfies all these conditions. What would be the number of coins left if there were 50 friends?

a) 18

b) 28

c) 8

d) 38

6) Answer (C)

Solution:

Let the total number of gold coins be ‘n’. We have been given that ‘n’ is of the form 20k + 18 = 19m + 17 + 18n + 16 = 15l + 13 . . . . . .
We can see that the terms are leaving a remainder of -2 on division with the different divisors. Hence, the required number of gold coins will be
LCM(1 to 20) – 2
LCM of first 20 numbers will be
Highest power of 2 among numbers from 1 to 20 = $2^4$
Highest power of 3 among numbers from 1 to 20 = $3^2$
Highest power of 5 among numbers from 1 to 20 = 5
Hence, the LCM will be
16*9*5*7*11*13*17*19
Hence, the required number will be 16*9*5*7*11*13*17*19 – 2
Now we need to find the remainder when this number is divided by 50.
{(16*9)(11*5)(7*13)(17*19) – 2} mod 50
=> {-6*5*-9*23 – 2} mod 50
=> (20*23 – 2) mod 50
=> 458 mod 50 = 8
Hence, the required answer is 8.

Question 7: What is the remainder when $1781^{1170}$ is divided by 385?

a) 384

b) 1

c) 2

d) 277

7) Answer (B)

Solution:

385 = 5 X 7 X 11
When $1781^{1170}$ is divided by 5, it leaves a remainder of $1^{1170}$=1. This means the number is in the form of 5l+1, where l is a positive integer.
Similarly, when $1781^{1170}$ is divided by 7, it leaves a remainder of $3^{1170}=27^{390}=(-1)^{390}$=1. This means the number is in the form of 7m+1, where m is a positive integer.
When $1781^{1170}$ is divided by 11, it leaves a remainder of $10^{1170}=(-1)^{1170}$=1. This means the number is in the form of 11n+1, where n is a positive integer.
Let the number be A. => A=5l+1=7m+1=11n+1
Taking the case of 5 and 7 we get, 5l=7m. First positive integral values of (l,m) is (7,5). This means that smallest value that satisfies the conditions 5l+1 and 7m+1 is 36, and all such values will be in the form of 35k+36.
Solving this with 11n+1, we get, 11n+1=35k+36. First positive integral values of (k,n) is (10,35). Thus smallest such value will be 11 X 35 + 1 or 385+1. Thus the number is in the form of 385p+385+1 which will leave a remainder of 1 when divided by 385. Thus, the answer is 1.

Question 8: Find the remainder when $30^{20} + 20^{30}$ is divided by 7

a) 5

b) 3

c) 4

d) 2

8) Answer (A)

Solution:

$30 ^{20}$mod 7 is the same as $2^{20}$ mod 7 which is 4.

Similarly

$20 ^{30}$ mod 7 is the same as ${-1} ^ {30}$ mod 7 = 1.

The remainder therefore is 4+1 =5.

Question 9: How many 6 digit numbers A exist such that
(i) The first and last digits of A are the same.
(ii) The sum of the digits of A is divisible by 3.
(iii) A is divisible by 5

a) 2000

b) 3333

c) 4000

d) 5000

9) Answer (B)

Solution:

Based on the first and the last statement, the number A is of the form 5abcd5.
We need to find the number of permutations of the digits a,b,c and d such that 10+a+b+c+d is divisible by 3.
This is equal to the permutations of the digits a,b,c and d such a+b+c+d leaves a remainder of 2 when divided by 3.
This is equal to the number of natural numbers less than 10000 which leave a remainder of 2 when divided by 3.
This equals 9999/3 = 3333

Hence, the correct answer is option (B)

Question 10: Find the remainder when $43564^{1^3 + 2^3 + 3^3 + …. + 27^3}$ is divided by 5.

10) Answer: 1

Solution:

The number 43564 ends in 4.
$4^1 \mod 5 = 4$
$4^2 \mod 5 = 1$
$4^3 \mod 5 = 4$
$4^4 \mod 5 = 1$
…
So, the remainder when odd powers of 4 are divided by 5 is 4 and the remainder when even powers of 4 are divided by 5 is 1.
$1^3 + 2^3 + 3^3 +…+27^3 = (27*28/2)^2= 27*14*27*14$ – even number which is a multiple of 4
So, the remainder is 1

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Question 11: A man covers half of his journey by train at 90 km/hr, one-third of the remainder by bus at 30 km/hr and the rest by cycle at 10 km/hr. The average speed during the entire journey is ______

a) 22.5 km/hr

b) 28.5 km/hr

c) 30.0 km/hr

d) 32.5 km/hr

11) Answer (A)

Solution:

Average speed = Total distance/total time

Let us consider total distance =180km

Case 1 :  man covers half of his journey by train at 90 km/hr

Distance= 90km => time t1=$\frac{90}{90}$=1hr

Remaining distance = 90km

Case 2 :  man covers one-third of the remainder by bus at 30 km/hr

Distance = 30km => time t2=$\frac{30}{30}$=1hr

Case 3 : man covers rest by cycle at 10 km/hr

Distance = 60km => time t3=$\frac{60}{10}$=6hr

Therefore Average speed = $\frac{180}{1+1+6}$= 22.5km/hr

Question 12: What is the remainder when $113^{24}$ is divided by 119?

a) 118

b) 1

c) 50

d) 49

12) Answer (C)

Solution:

113 mod 119 = -6
So $113^{24}$ mod 119 = $6^{24}$ mod 119
119 = 17*7
$6^{24}$ mod 7 = 1
Euler of 17 = 16. 6 and 17 are coprime. So, $6^{24}$ mod 17 = $6^8$ mod 17
$6^8$ mod 17 = $2^8$ mod 17 *$3^8$ mod 17 = 1 * 81 mod 17 * 81 mod 17 = 13*13 mod 17 = -1 = 16
Using Chinese Remainder Theorem,

17k + 16 = 7p + 1
This is satisfied for k = 2 and p = 7. Value of 17k + 15 = 50

So, the remainder = 50

Question 13: What is the remainder when $35^{49}$ is divided by 100?

a) 0

b) 25

c) 75

d) None of the above

13) Answer (C)

Solution:

The expression can be written as $(30 + 5)^{49}$
When expanded, the terms of this series are: $^{49}C_0\ 30^{49}*5^0 + ^{49}C_1\ 30^{48}*5^1 + …. + ^{49}C_{48}\ 30^1*5^{48} + ^{49}C_{49}\ 30^0*5^{49}$
When divided by 100, all the first 48 terms give a remainder of 0.
Last two terms: $49\ *\ 30\ *\ 5^{48}\ +\ 5^{49}\ =\ 5^{48}\ *\ (1470\ +\ 5)\ =\ 1475\ *\ 5^{48}$

We have to find $1475\ *\ 5^{48}$ mod 100
Powers of 5 always end in 25
So, the expression boils down to: 1475 mod 100 * 25 = 75 * 25 mod 100 = 3 * 625 mod 100 = 3 * 25 = 75

Question 14: Find the sum of the last two digits of 1! + 4! + 9! + 16! + 25! . . . . . . . + 961!.

a) 3

b) 5

c) 6

d) 8

14) Answer (B)

Solution:

Last two digits are nothing but remainder when the number is divided by 100.
All the terms after 10! will be divisible by 100. So we need to find the remainder of first 3 terms with 100.
So required remainder is
1 + 24 + 80 = 105
=> 105 mod 100 = 05
So last two digits are 05.
Hence the sum is 5.

Question 15: Find the unit digit of $367^{5^{763}}$?

a) 1

b) 3

c) 7

d) 9

15) Answer (C)

Solution:

Any $5^n$ where n>1 will end with the digits 25. As numbers ending in 24 are divisible by 4, any $5^n$ where n>1 can be represented in the form of 4k+1.

To find the last digit, we need to find the remainder when the number is divided by 10.

We split 367 as 360+7. In the expansion of $(360+7)^{5^{763}}$ all multiples of 360 would be divisible by 10.

Hence, $367^{5^{763}}$ mod 10 = $7^{5^{763}}$ mod 10 =$7^{4k+1}$ mod 10

7 mod 10 = 7

$7^2$ mod 10 = 9

$7^3$ mod 10 = 3

$7^4$ mod 10 = 1

$7^5$ mod 10 = 7 and so on. Hence, the last digit of powers of 7 have a cyclicity of 4.

Thus, $7^{4k+1}$ mod 10 = 7

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Question 16: What is the remainder when 123412341234………(1234 digits) is divided by 625?

16) Answer: 287

Solution:

625 = $5^4$, so we have to check the remainder obtained when the last 4 digits are divided by 625
12341234……..123412
3412 is the last 4 digits
The remainder obtained when 3412 is divided by 625 is 287.
Hence, 287 is the correct answer.

Question 17: A number ‘n’ when divided by 6 leaves a remainder of ‘k’ and when divided by 12 leaves a remainder of ‘3k’. How many values can ‘n’ take if it is known that ‘n’ and ‘k’ are natural numbers less than 100?

17) Answer: 8

Solution:

It has been given that the number leaves a remainder of ‘k’ when divided by 6 and 3k when divided by 12.
Let the number be 12x + 3k.
Now, 12x will always be divisible by 6. Therefore, out of 3k, 2k should be divisible by 6 for the number to leave a remainder of ‘k’ when divided by 6.
=> 2k should be equal to 6.
=> k = 3.
Now, we know that the number is of the form 12x+9.
x can vary from 0 to 7.
Therefore, x (and ‘n’) can take a total of 8 values.

Question 18: What is the remainder when $77^{59}+33^{59}$ is divided by 220?

a) 0

b) 55

c) 110

d) 219

e) 1

18) Answer (C)

Solution:

We can divide the divisor in two co-prime factor = 220 = 4*55

Case 1: The remainder when $77^{59}+33^{59}$ is divided by 55.
We know that $x^n + y^n$ is divisible by (x+y) if n is an odd number. Hence, we can say that $77^{59}+33^{59}$ is divisible by 110. So we can say that $77^{59}+33^{59}$ is divisible by 55 as well. Therefore, $77^{59}+33^{59}$ is a 55k type number where k is an integer.

Case 2: The remainder when $77^{59}+33^{59}$ is divided by 4.
Rem. ($\dfrac{77^{59}+33^{59}}{4}$) = Rem. ($\dfrac{(76+1)^{59}+(32+1)^{59}}{4}$) = 2.
Hence, we can say that $77^{59}+33^{59}$ is a 4m+2 type of number.

We have to find the least positive integral value of k and m such that
55k = 4m + 2
We can see that k = 2 and m = 27 is a possible solution.
Hence, we can say that remainder = 55*2 = 110.
Therefore, option C is the correct answer.

Alternate solution:

Let a=77;

then 110-a=33

writing  $77^{59}+33^{59}$ in terms of a;

=$a^{59}+(110-a)^{59}$

applying binomial expansion;

=$a^{59}+\left(110^{59}+59_{C_1}110^{58}\left(-a\right)^1+59_{C_2}110^{57}\left(-a\right)^2+…..59_{C_{58}}110^{ }\left(-a\right)^{58}+\left(-a\right)^{59}\right)$

=$110^{59}+59_{C_1}110^{58}\left(-a\right)^1+59_{C_2}110^{57}\left(-a\right)^2+…..59_{C_{58}}110^{ }\left(-a\right)^{58}$

Also, 110 = 220/2

so $110^2\ will\ be\ divisible\ by\ 220$

So all the term having power of 110 more than or equal to 2 are divisible by 220.

Only penultimate term of expansion have power of 110 less than 2; so we will get remainder by these terms only.

Rewriting the term : $59\cdot110\cdot\left(-a\right)^{58}$ mod 220

$59\cdot110\cdot\left(-77\right)^{58}$

Now there are only 2 remainder possible 0 or 110;

by checking last term of$59\cdot\left(-77\right)^{58}$ we find it is odd which will give a remainder of 110.

Question 19: What is the sum of all multiples of 3 less than 1000 which give an odd remainder when divided by 11 ?

19) Answer: 75837

Solution:

The LCM of 3 and 11 is 33. Thus, the number of numbers we find in the first 33 numbers will be the number of numbers that are there in every consecutive set 33 numbers.

Multiples of 3 less than 33 that have odd remainders when divided by 11 : 3,9,12,18,27

In the next set of 33 numbers, the numbers that satisfy the conditions are : 36,42,45,51,60 ie 33+3,33+9,33+12,33+18,33+27 respectively.

Below 1000, there will be $\dfrac{1000}{33} = 30 \dfrac{10}{33}$ which means there will be 30 such sets.

From 33×30 to 1000, ie from 990 to 1000, there are only 2 more numbers that satisfy the conditions ie 993 and 999.

To find sum of all the numbers, first we find the sum of the first series of numbers ie all numbers below 33 that satisfy the condition.

$S_1 = 3+9+12+18+27 = 69 $

The sum of the second series of numbers $S_2$ = (3+33)+(9+33)+(12+33)+(18+33)+(27+33) = 69 + (5 * 33) = 69+165

Similarly, the sum of the third series of numbers $S_3 = 69+(165 \times 2)$

Thus, sum of the $n^{th}$ series of numbers $S_n = 69+[165 \times (n-1)] $

Thus, total sum of the series till the $30^{th}$ set of numbers $S = (69 \times 30) + [165 \times (1+2+3…..29)] $

$ \Rightarrow S = 2070 + [165 \times \dfrac{29 \times 30}{2}] $

$ \Rightarrow S = 2070 + 71775 = 73845 $

We should remember that this series does not take into consideration the last 2 numbers.

Therefore the actual sum $S’ = 73845 + 993 + 999 = 75837 $

Question 20: What is the remainder when 1*1! + 2*2! + 3*3! + . . . + 19*19! +20*20! is divided by 21

a) 0

b) 1

c) 19

d) 20

20) Answer (D)

Solution:

We know n! = 1*2*3…….*(n-1)(n)
Thus, 7! = 1*2*3…..*7
Thus, any factorial greater than 7 will be multiple of 21.
Hence, remainder when divided by 21 of any factorial greater than 7 will be 0.
Thus, we need to find the remainder of 1*1+2*2!+3*3!+4*4!+5*5!+6*6! when divided by 21
The sum of 1*1+2*2!+3*3!+4*4!+5*5!+6*6! will be 5039 and reminder when divided by 21 will be 20.
Thus, option D is the correct answer.

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