# All Number Systems Questions (CAT 2017-22) | CAT PYQs [Download PDF]

Number Systems is one of the key topics in the CAT Quantitative Ability (QA) Section. It is essential that you know the **basics of the CAT Number Systems** well and practice the questions. Also, do check out all the Number System questions from the **CAT Previous Papers** with detailed video solutions. This article will look into some Number System Questions for the CAT Exam. If you want to practice these important Number System questions, you can download the PDF, which is completely Free.

If you’re a CAT aspirant gearing up for one of the most competitive management entrance tests in India, you understand the significance of acing each section. In this blog, we delve into the heart of CAT preparation, focusing specifically on the Number Systems segment. We will walk you through the previous year’s questions from 2017 to 2022, offering step-by-step solutions and valuable insights to equip you for success.

Number Systems is a critical component of the CAT exam, and tackling it with precision is a must. Whether new to CAT or a seasoned CAT warrior, this blog is your one-stop destination for honing your skills and boosting your confidence. We’ve gathered past year questions, expert tips, and strategies to help you excel in this challenging section.

Let’s embark on this CAT 2023 preparation journey, dissecting Number Systems and arming you with the knowledge and techniques you need to conquer the CAT exam.

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**Question 1:Â **If $a, b, c,$ and $d$ are integers such that $a+b+c+d=30$ then the minimum possible value of $(a – b)^{2} + (a – c)^{2} + (a – d)^{2}$Â is

**1)Â Answer:Â 2**

**Solution:**

For the value of given expression to be minimum, the values of $a, b, c$ and $d$ should be as close as possible. 30/4 = 7.5. Since each one of these are integers so values must be 8, 8, 7, 7. On putting these values in the given expression, we get

$(8 – 8)^{2} + (8 – 7)^{2} + (8 – 7)^{2}$

=> 1 + 1 = 2

**Question 2:Â **If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

a)Â 1777

b)Â 1785

c)Â 1875

d)Â 1877

**2)Â AnswerÂ (D)**

**Solution:**

$(x -1)x(x+1) = 15600$

=> $x^3 – x= 15600 $

The nearest cube to 15600 is 15625 = $25^3$

We can verify that x = 25 satisfies the equation above.

Hence the three numbers are 24, 25, 26. Sum of their squares = 1877

**Question 3:Â **While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

**3)Â Answer:Â 40**

**Solution:**

We know that one of the 3 numbers is 37.

Let the product of the other 2 numbers be x.

It has been given that 73x-37x = 720

36x = 720

x = 20

Product of 2 real numbers is 20.

We have to find the minimum possible value of the sum of the squares of the 2 numbers.

Let x=a*b

It has been given that a*b=20

The least possible sum for a given product is obtained when the numbers are as close to each other as possible.

Therefore, when a=b, the value of a and b will be $\sqrt{20}$.

Sum of the squares of the 2 numbers = 20 + 20 = 40.

Therefore, 40 is the correct answer.

**Question 4:Â **The number of integers x such that $0.25 \leq 2^x \leq 200$ and $2^x + 2$ is perfectly divisible by either 3 or 4, is

**4)Â Answer:Â 5**

**Solution:**

At **$x = 0, 2^x = 1$** which is in the given range [0.25, 200]

$2^x + 2$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution.

At **$x = 1, 2^x = 2$** which is in the given range [0.25, 200]

$2^x + 2$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution.

At **$x = 2, 2^x = 4$** which is in the given range [0.25, 200]

$2^x + 2$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution.

At **$x = 3, 2^x = 8$** which is in the given range [0.25, 200]

$2^x + 2$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can’t be a solution.

At **$x = 4, 2^x = 16$** which is in the given range [0.25, 200]

$2^x + 2$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution.

At **$x = 5, 2^x = 32$** which is in the given range [0.25, 200]

$2^x + 2$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can’t be a solution.

At **$x = 6, 2^x = 64$** which is in the given range [0.25, 200]

$2^x + 2$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution.

At **$x = 7, 2^x = 128$** which is in the given range [0.25, 200]

$2^x + 2$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can’t be a solution.

At **$x = 8, 2^x = 256$** which is not in the given range [0.25, 200]. Hence, x can’t take any value greater than 7.

Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that ‘x’ can take 5 different integer values.

**Question 5:Â **If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

a)Â -32

b)Â 16

c)Â 48

d)Â 64

**5)Â AnswerÂ (D)**

**Solution:**

Let ‘a’ and ‘b’ are those two numbers.

$\Rightarrow$ $a^2+b^2 = 97$

$\Rightarrow$ $a^2+b^2-2ab = 97-2ab$

$\Rightarrow$ $(a-b)^2 = 97-2ab$

We know that $(a-b)^2$ $\geq$ 0

$\Rightarrow$ 97-2ab $\geq$ 0

$\Rightarrow$ ab $\leq$ 48.5

Hence, ab $\neq$ 64. Therefore, option D is the correct answer.

**Question 6:Â **The smallest integer n for which $4^{n} > 17^{19}$ holds, is closest to

a)Â 37

b)Â 35

c)Â 33

d)Â 39

**6)Â AnswerÂ (D)**

**Solution:**

$4^{n} > 17^{19}$

$\Rightarrow$ $16^{n/2} > 17^{19}$

Therefore, we can say thatÂ n/2 >Â 19

n > 38

Hence, option D is the correct answer.

**Question 7:Â **What is the largest positive integer n such that $\frac{n^2 + 7n + 12}{n^2 – n – 12}$ is also a positive integer?

a)Â 6

b)Â 16

c)Â 8

d)Â 12

**7)Â AnswerÂ (D)**

**Solution:**

$\ \frac{\ n^2+3n+4n+12}{n^2-4n+3n-12}$

=$\ \frac{\ n^{ }\left(n+3\right)+4\left(n+3\right)}{n^{ }\left(n-4\right)+3\left(n-4\right)}$

=$\ \frac{\left(\ n+4\right)\left(n+3\right)}{\left(n-4\right)\left(n+3\right)}$

=$\ \frac{\left(\ n+4\right)}{\left(n-4\right)}$

=$\ \frac{\left(\ n-4\right)+8}{\left(n-4\right)}$

=$\ 1+\frac{8}{\left(n-4\right)}$ which will be maximum when n-4 =8

n=12

D is the correct answer.

**Question 8:Â **How many pairs (m, n) of positive integers satisfy the equation $m^2 + 105 = n^2$?

**8)Â Answer:Â 4**

**Solution:**

$n^2-m^2=105$

(n-m)(n+m) = 1*105, 3*35, 5*21, 7*15, 15*7, 21*5, 35*3, 105*1.

n-m=1, n+m=105Â ==> n=53, m=52

n-m=3, n+m=35 ==> n=19, m=16

n-m=5, n+m=21Â ==> n=13, m=8

n-m=7, n+m=15 ==> n=11, m=4

n-m=15, n+m=7 ==> n=11, m=-4

n-m=21, n+m=5 ==> n=13, m=-8

n-m=35, n+m=3 ==> n=19, m=-16

n-m=105, n+m=1 ==> n=53, m=-52

Since only positive integer values of m and n are required. There are 4 possible solutions.

**Question 9:Â **How many factors of $2^4 \times 3^5 \times 10^4$ are perfect squares which are greater than 1?

**9)Â Answer:Â 44**

**Solution:**

$2^4 \times 3^5 \times 10^4$

=$2^4 \times 3^5 \times 2^4*5^4$

=$2^8 \times 3^5 \times 5^4$

For the factor to be a perfect square, the factor should be even power of the number.

In $2^8$, the factors which are perfect squares are $2^0, 2^2, 2^4, 2^6, 2^8$ = 5

Similarly, in $3^5$, the factors which are perfect squares areÂ $3^0, 3^2, 3^4$ = 3

In $5^4$,Â the factors which are perfect squares are $5^0, 5^2, 5^4$ = 3

Number of perfect squares greater than 1 = 5*3*3-1

=44

**Question 10:Â **In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum ofÂ fifth and sixth digits. Then, the largest possible value of the fourth digit is

**10)Â Answer:Â 7**

**Solution:**

Let the six-digit number be ABCDEF

F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.

Therefore D = 2A+2B+C = 2A + 4A + A= 7A.

A cannot be 0 as the number is a 6 digit number.

A cannot be 2 as D would become 2 digit number.

Therefore A is 1 and D is 7.

**Question 11:Â **The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

a)Â 58

b)Â 85

c)Â 50

d)Â 95

**11)Â AnswerÂ (C)**

**Solution:**

Assume the numbers are a and b, then ab=616

We have,Â $\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$ = $\ \frac{\ 157}{3}$

=>Â $\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$

=> $154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$Â = 0

=>Â $154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$ = 0Â Â Â Â Â (ab=616)

=>$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$Â Â (154*4=616)

=>Â $\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$

=>Â $a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$

Adding ab=616 on both sides, we get

$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$

=>Â $\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$ = 2500

=> a+b=50

**Question 12:Â **How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

**12)Â Answer:Â 21**

**Solution:**

Let the number be ‘abc’. Then, $2<a\times\ b\times\ c<7$. The product can be 3,4,5,6.

We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of **12** numbers fulfilling the criteria.

We can factories 4 as 2*2 and the combination 2,2,1 can be used to form **3** more distinct numbers.

We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form **6** additional distinct numbers.

Thus a total of 12 + 3 + 6 = **21** such numbers can be formed.

**Question 13:Â **The mean of all 4-digit even natural numbers of the form ‘aabb’,where $a>0$, is

a)Â 4466

b)Â 5050

c)Â 4864

d)Â 5544

**13)Â AnswerÂ (D)**

**Solution:**

The four digit even numbers will be of form:

1100, 1122, 1144 … 1188, 2200, 2222, 2244 … 9900, 9922, 9944, 9966, 9988

Their sum ‘S’ will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+…)….+(9900+9900+22+9900+44+9900+66+9900+88)

=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)….+9900*5+(22+44+66+88)

=> S=5*1100(1+2+3+…9)+9(22+44+66+88)

=>S=5*1100*9*10/2 + 9*11*20

Total number of numbers are 9*5=45

.’. Mean will be S/45 = 5*1100+44=5544.

Option D

**Question 14:Â **If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is

a)Â 49

b)Â 56

c)Â 59

d)Â 46

**14)Â AnswerÂ (D)**

**Solution:**

Since $c<9$, we can have the following viable combinations forÂ $b\times\ c\ =96$ (given our objective is to minimize the sum):

$48\times\ 2$ ;Â $32\times3$ ;Â $24\times\ 4$ ;Â $16\times6$ ;Â $12\times8$

Similarly, we can factorizeÂ $a\times\ b\ = 432$ into its factors. On close observation, we notice thatÂ $18\times24\ and\ 24\ \times\ 4\ $ corresponding toÂ $a\times b\ and\ b\times\ c\ $ respectivelyÂ together render us with the least value of the sum ofÂ $a+b\ +\ c\ \ =\ 18+24+4\ =46$

Hence, Option D is the correct answer.

**Question 15:Â **Let m and n be natural numbers such that n is even and $0.2<\frac{m}{20},\frac{n}{m},\frac{n}{11}<0.5$. Then $m-2n$ equals

a)Â 3

b)Â 1

c)Â 2

d)Â 4

**15)Â AnswerÂ (B)**

**Solution:**

$0.2<\frac{n}{11}<0.5$

=> 2.2<n<5.5

Since n is an even natural number, the value of n = 4

$0.2<\frac{m}{20}<0.5$Â => 4< m<10. Possible values of m = 5,6,7,8,9

SinceÂ $0.2<\frac{n}{m}<0.5$, the only possible value of m is 9

Hence m-2n = 9-8 = 1

**Question 16:Â **How many integers in the set {100, 101, 102, …, 999} have at least one digitÂ repeated?

**16)Â Answer:Â 252**

**Solution:**

Total number of numbers from 100 to 999 = 900

The number of three digits numbers with unique digits:

_ _ _

The hundredth’s place can be filled in 9 ways ( Number 0 cannot be selected)

Ten’s place can be filled in 9 ways

One’s place can be filled in 8 ways

Total number of numbers = 9*9*8 = 648

Number of integers in the set {100, 101, 102, …, 999} have at least one digit repeated = 900 – 648 = 252

**Question 17:Â **Let N, x and y be positive integers such that $N=x+y,2<x<10$ and $14<y<23$. If $N>25$, then how many distinct values are possible forÂ N?

**17)Â Answer:Â 6**

**Solution:**

Possible values of x = 3,4,5,6,7,8,9

When x = 3, there is no possible value of y

When x = 4, the possible values of y = 22

When x = 5, the possible values of y=21,22

When x = 6, the possible values of y = 20.21,22

When x = 7, the possible values of y = 19,20,21,22

When x = 8, the possible values of y=18,19,20,21,22

When x = 9, the possible values of y=17,18,19,20,21,22

The unique values of N = 26,27,28,29,30,31

**Question 18:Â **How many of the integers 1, 2, â€¦ , 120, are divisible by none of 2, 5 and 7?

a)Â 42

b)Â 41

c)Â 40

d)Â 43

**18)Â AnswerÂ (B)**

**Solution:**

The number of multiples of 2 between 1 and 120 = 60

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7

Hence,Â numberÂ of the integers 1, 2, â€¦ , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41

**Question 19:Â **How many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ ?

a)Â 2018

b)Â 2019

c)Â 2017

d)Â 2020

**19)Â AnswerÂ (A)**

**Solution:**

$ab\ =\ 4^{2017}=2^{4034}$

The total number of factors = 4035.

out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

And since the given number is a perfect square we have one set of two equal factors.

.’.Â many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ = 2018.

**Question 20:Â **How many 4-digit numbers, each greater than 1000 and each having all four digitsÂ distinct, are there with 7 coming before 3?

**20)Â Answer:Â 315**

**Solution:**

Here there are two cases possible

Case 1: When 7 is at the left extreme

In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)

So total ways 3(8)(7)= 168

Case 2: When 7 is not at the extremes

Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left)

Hence in total 3(7)(7)=147 ways

Total ways 168+147=315 ways

**Question 21:Â **For all possible integers n satisfying $2.25\leq2+2^{n+2}\leq202$, then the number of integer values of $3+3^{n+1}$ is:

**21)Â Answer:Â 7**

**Solution:**

$2.25\leq2+2^{n+2}\leq202$

$2.25-2\le2+2^{n+2}-2\le202-2$

$0.25\le2^{n+2}\le200$

$\log_20.25\le n+2\le\log_2200$

$-2\le n+2\le7.xx$

$-4\le n\le7.xx-2$

$-4\le n\le5.xx$

Possible integers = -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

If we see the second expression that is provided, i.e

$3+3^{n+1}$, it can be implied that n should be at least -1 for this expression to be an integer.

So, n = -1, 0, 1, 2, 3, 4, 5.

Hence, there are a total of 7 values.

**Question 22:Â **For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

**22)Â Answer:Â 4195**

**Solution:**

Given the 4 digit number :

Considering the number in thousands digit is a number in the hundredth digit is b, number in tens digit is c, number in the units digit is d.

Let the number be abcd.

Given that a+b+c = 14. (1)

b+c+d = 15. (2)

c = d+4. (3).

In order to find the maximum number which satisfies the condition, we need to have abcd such that a is maximum which is the digit in thousands placeÂ in order to maximize the value of the number. b, c, and d are less than 9 each as they are single-digit numbers.

Substituting (3) in (2) we have b+d+4+d = 15, b+2*d = 11.Â (4)

Subtracting (2) and (1) : (2) – (1) = d = a+1.Â Â (5)

Since c cannot be greater than 9 considering c to be the maximum valueÂ 9 the value of d is 5.

If d = 5, using d = a+1, a = 4.

Hence the maximum value of a = 4 when c = 9, d = 5.

Substituting b+2*d = 11. b = 1.

The highest four-digit number satisfying the conditionÂ is 4195

**Question 23:Â **Let A be the largest positive integer that divides all the numbers of the form $3^k + 4^k + 5^k$, and B be the largest positive integer that divides all the numbers of the form $4^k + 3(4^k) + 4^{k + 2}$ , where k is any positive integer. Then (A + B) equals

**23)Â Answer:Â 82**

**Solution:**

A is the HCF ofÂ $3^k+4^k+5^k$ for different values of k.

For k = 1, value is 12

For k = 2, value is 50

For k = 3, value is 216

HCF is 2. Therefore, A = 2

$4^k+3\left(4^k\right)+4^{k+2}=4^k\left(1+3\right)+4^{k+2}=4^{k+1}+4^{k+2}=4^{k+1}\left(1+4\right)=5\cdot4^{k+1}$

HCF of the values isÂ when k = 1, i.e. 5*16 = 80

Therefore, B = 80

A + B = 82

**Question 24:Â **For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is

a)Â 4

b)Â 7

c)Â 6

d)Â 5

**24)Â AnswerÂ (B)**

**Solution:**

To find the largest possible value of n, we need to find the value of n such that n! is less than 15000.

7! = 5040

8! = 40320 > 15000

This implies 15000! is not divisible by 40320!

Therefore, maximum value n can take is 7.

The answer is option B.

**Question 25:Â **Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is

a)Â 23

b)Â 24

c)Â 23.5

d)Â 22.5

**25)Â AnswerÂ (D)**

**Solution:**

Let the six numbers be a, b, c, d, e, f in ascending order

a+b = 28

e+f =Â 56

If we want to maximise the average then we have to maximise the value of c and d and maximise e and minimise f

e+f = 56

As e and f are distinct natural numbers so possible values are 27 and 29

Therefore c and d will be 25 and 26 respecitively

So average =Â $\frac{\left(a+b+c+d+e+f\right)}{6}=\frac{\left(28+25+26+56\right)}{6}=\frac{135}{6}=22.5$

**Question 26:Â **A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

**26)Â Answer:Â 150**

**Solution:**

Since the total number of students, when divided byÂ either 9 or 10 or 12 or 25 each, gives a remainder of 4, the number will be in the form of LCM(9,10,12,25)k + 4 = 900k + 4.

It is given that the value of 900k + 4 is less than 5000.

Also, it is given that 900k + 4 is divided by 11.

It is only possible when k = 2 and total students = 1804.

So, the number of 12 students group = 1800/12 = 150.