# CAT Number System Questions PDF [Most Important]

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Number System for CAT is one of the key topics in the CAT Quants section. If you’re weak in Number Systems, make sure you are at least aware of the basics. Learn all the important formulas in Number systems. You can check out these CAT Number System questions from the CAT Previous year’s papers. Also, if you don’t have enough time, learn only the basics of Number Systems and practice a few easy questions from the topic. This post will look at important Number System questions in the CAT quant section. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this CAT Number System Questions PDF along with the detailed solutions (and video solutions) below, which is completely Free.

Question 1:Â What are the last two digits of $7^{2008}$?

a)Â 21

b)Â 61

c)Â 01

d)Â 41

e)Â 81

Solution:

$7^4$ = 2401 = 2400+1
So, any multiple of $7^4$ will always end in 01
Since 2008 is a multiple of 4, $7^{2008}$ will also end in 01

Question 2:Â A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

a)Â $2 \leq x \leq 6$

b)Â $5 \leq x \leq 8$

c)Â $9 \leq x \leq 12$

d)Â $11 \leq x \leq 14$

e)Â $13 \leq x \leq 18$

Solution:

After the first sale, the remaining quantity would be (x/2)-0.5 and after the second sale, the remaining quantity is 0.25x-0.75

After the last sale, the remaining quantity is 0.125x-(7/8) which will be equal to 0

SoÂ 0.125x-(7/8) = 0 => x = 7

Question 3:Â How many even integers n, where $100 \leq n \leq 200$ , are divisible neither by seven nor by nine?

a)Â 40

b)Â 37

c)Â 39

d)Â 38

Solution:

Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 – (7+6-1) = 39

There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.

Question 4:Â The number of positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*â€¦*3*2*1 is not divisible by n is

a)Â 5

b)Â 7

c)Â 13

d)Â 14

Solution:

positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*â€¦*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.

Question 5:Â Let T be the set of integers {3,11,19,27,â€¦451,459,467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

a)Â 32

b)Â 28

c)Â 29

d)Â 30

Solution:

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.

Now S will have atleast have of 59 terms i.e 29 .

Also the sum of 29th term and 30th term is less than 470.

Hence, maximum possible elements in S is 30.

Question 6:Â The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?

a)Â 21

b)Â 25

c)Â 41

d)Â 67

e)Â 73

Solution:

Maximum sum of the four numbers <= 384=99+97+95+93
384/10 = 38.4
So, the perfect square is a number less than 38.4
The possibilities are 36, 25, 16 and 9
For the sum to be 360, the numbers can be 87, 89, 91 and 93
The sum of four consecutive odd numbers cannot be 250
For the sum to be 160, the numbers can be 37,39,41 and 43
The sum of 4 consecutive odd numbers cannot be 90
So, from the options, the answer is 41.

Question 7:Â For a positive integer n, let $P_n$ denote the product of the digits of n, and $S_n$ denote the sum of the digits of n. The number of integers between 10 and 1000 for which $P_n$ + $S_n$ = n is

a)Â 81

b)Â 16

c)Â 18

d)Â 9

Solution:

Let n can be a 2 digit or a 3 digit number.

First letÂ n be a 2 digit number.
So n = 10x + y and Pn = xy and Sn = x + y
Now, Pn + Sn = n
Therefore, xy + x + y = 10x + y , we have y = 9 .
Hence there are 9 numbers 19, 29,..Â ,99, so 9 cases .

Now ifÂ n isÂ a 3 digit number.
Let n = 100x + 10y + z
So Pn = xyz and Sn = x + y + z
Now, for Pn + Sn = n ; Â xyz + x + y + z = 100x + 10y + z ; so.Â xyz = 99x + 9y .
For above equation there is no value for which the above equation have an integer (singleÂ digit) value.

Hence option D.

Question 8:Â Let S be a set of positive integers such that every element n of S satisfies the conditions
A. 1000 <= n <= 1200
B. every digit in n is odd
Then how many elements of S are divisible by 3?

a)Â 9

b)Â 10

c)Â 11

d)Â 12

Solution:

The no. has all the digits as odd no. and is divisible by 3. So the possibilities are

1113
1119
1131
1137
1155
1173
1179
1191
1197
Hence 9 possibilities .

Question 9:Â Of 128 boxes of oranges, each box contains at least 120 and at most 144 oranges. X is the maximum number of boxes containing the same number of oranges. What is the minimum value of X?

a)Â 5

b)Â 103

c)Â 6

d)Â Cannot be determined

Solution:

Each box contains at least 120 and at most 144 oranges.

So boxes may contain 25 different numbers of oranges among 120, 121, 122, …. 144.

Lets start counting.

1st 25 boxes contain different numbers of oranges and this is repeated till 5 sets as 25*5=125.

Now we have accounted for 125 boxes. Still 3 boxes are remaining. These 3 boxes can have any number of oranges from 120 to 144.

Already every number is in 5 boxes. Even if these 3 boxes have different number of oranges, some number of oranges will be in 6 boxes.

Hence the number of boxes containing the same number of oranges is at least 6.

Question 10:Â Let n be the number of different five-digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

a)Â 144

b)Â 168

c)Â 192

d)Â None of these

Solution:

To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4 . So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number. Remaining 3 digits out of 4 can be selected in $^4C_3$ ways and further can be arranged in 3! ways . So in total = 8*4*6 = 192

Question 11: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

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 12: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

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 13: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

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

Question 14:Â How many pairs (m, n) of positive integers satisfy the equation $m^2 + 105 = n^2$?

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

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 16: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

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 17: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

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

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 19: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

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 20: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

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 $a+b\ +\ c\ \ =\ 18+24+4\ =46$