# CAT Number System Questions (with Notes) PDF

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Number Systems is one of the key topics in the CAT Quants section. Over the past few years, CAT Number System questions have made a recurrent appearance in the Quants section. You can expect around 1-2 questions in the 22-question format of the CAT Quant section. You can check out these  CAT Number System Questions from Previous years. In this article, we will look into some important CAT Number System Questions (with Notes) PDF. These are a good source for practice; If you want to practice these questions, you can download these CAT Number System Questions PDF below, which is completely Free.

• CAT Number System Questions – Tip 1: A few CAT Number System questions appear in the CAT and other MBA entrance exams every year. If you’re starting the prep, firstly understand the CAT Number System Syllabus; Based on our analysis of the previous years CAT number system questions,  this was number system weightage in CAT: 1 question was asked on this topic (in CAT 2021).
• CAT Number System Questions – Tip 2: If you’re not very strong in this topic, learn at least the basics, and number systems cat tricks, so that if an easy question comes you will be able to answer it. Number System questions for CAT can be tackled with a strong foundation in this topic. Practice these CAT Number System questions PDF with solutions. Learn all the major formulae from these concepts. You can check out the CAT Number System questions and answers PDF and learn all the Important Number System for CAT tricks Formulas PDF here.

Let us now look at some very important CAT Number System Questions along with detailed video solutions.

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

a) 21

b) 61

c) 01

d) 41

e) 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: 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) $2 \leq x \leq 6$

b) $5 \leq x \leq 8$

c) $9 \leq x \leq 12$

d) $11 \leq x \leq 14$

e) $13 \leq x \leq 18$

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

So 0.125x-(7/8) = 0 => x = 7

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

a) 40

b) 37

c) 39

d) 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: The number of positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*…*3*2*1 is not divisible by n is

a) 5

b) 7

c) 13

d) 14

Solution:

positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*…*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: Let T be the set of integers {3,11,19,27,…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) 32

b) 28

c) 29

d) 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: 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) 21

b) 25

c) 41

d) 67

e) 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: The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?

a) 100<A<299

b) 106<A<305

c) 112<A<311

d) 118<A<317

Solution:

Let A = 100x + 10y + z  and B = 100z + 10y + x .According to given condition B – A = 99(z – x) As (B – A) is divisible by 7 . So clearly  (z – x) should be  divisible by 7.  z and x can have values 8,1 or 9,2 , such that 8-2=9-2=7 and  y can have  value from 0 to 9.
So Lowest possible value of A lowest x,y and z which is  is 108 and the highest possible value of A is 299.

Question 8: 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) 81

b) 16

c) 18

d) 9

Solution:

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

First let 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,.. ,99, so 9 cases .

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

Hence option D.

Question 9: 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) 9

b) 10

c) 11

d) 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 10: 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) 5

b) 103

c) 6

d) 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 11: 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) 144

b) 168

c) 192

d) 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 12: Let D be recurring decimal of the form, $D = 0.a_1a_2a_1a_2a_1a_2…$, where digits $a_1$ and $a_2$ lie between 0 and 9. Further, at most one of them is zero. Then which of the following numbers necessarily produces an integer, when multiplied by D?

a) 18

b) 108

c) 198

d) 288

Solution:

Case 1: $a_1=0$
So, D equals $0.0a_20a_20a_2…$
So, 100D equals $a_2.0a_20a_2…$
So, 99D equals $a_2$

Case 2: $a_2=0$
So, D equals $0.a_10a_10a_1…$
So, 100D equals $a_10.a_10a_1….$
So, 99D equals $a_10$

So, in both the cases, 99D is an integer. From the given options, only option C satisfies this condition (198=2*99) and hence the correct answer is C.

Question 13: If $x^2 + y^2 = 0.1$ and |x-y|=0.2, then |x|+|y| is equal to:

a) 0.3

b) 0.4

c) 0.2

d) 0.6

Solution:

$(x – y)^2 = x^2 + y^2 – 2xy$

$0.04 = 0.1 – 2xy => xy = 0.03$

So, |xy| = 0.03

$(|x| + |y|)^2 = x^2 + y^2 + 2|xy| = 0.1 + 0.06 = 0.16$

So, |x|+|y| = 0.4

Question 14: What is the greatest power of 5 which can divide 80! exactly?

a) 16

b) 20

c) 19

d) None of these

Solution:

The highest power of 5 in 80! = [80/5] + [$80/5^2$] = 16 + 3 = 19

So, the highest power of 5 which divides 80! exactly = 19

Question 15: If x is a positive integer such that 2x +12 is perfectly divisible by x, then the number of possible values of x is

a) 2

b) 5

c) 6

d) 12

Solution:

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)

Question 16: If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be

a) 7 and 8

b) 8 and 0

c) 5 and 8

d) None of these

Solution:

According to the divisible rule of 9, the sum of all digits should be divisible by 9.
i.e. 55+A+B = 9k
So sum can be either 63 or 72.
For 63, A+B should be 8.
In given options, option B has values of A and B whose sum is 8 and by putting them we are having a number which divisible by both 9 and 8.

Question 17: If n is an integer, how many values of n will give an integral value of $\frac{(16n^2+ 7n+6)}{n}$ ?

a) 2

b) 3

c) 4

d) None of these

Solution:

Expression can be reduced to 16n + 7 + $\frac{6}{n}$
Now to make above value  an integer n can be 1,2,3,6,-1,-2,-3,-6

Question 18: $n^3$ is odd. Which of the following statement(s) is/are true?
I. $n$ is odd.
II.$n^2$ is odd.
III.$n^2$ is even.

a) I only

b) II only

c) I and II

d) I and III

Solution:

if $n^3$ is odd then $n$ will be odd. let’s say it is $2k+1$
then $n^2$ will be = $(4k^2 + 4k + 1)$ which will be odd

Question 19: How many five digit numbers can be formed from 1, 2, 3, 4, 5, without repetition, when the digit at the unit’s place must be greater than that in the ten’s place?

a) 54

b) 60

c) 17

d) 2 × 4!

Solution:

Possible numbers with unit’s place as 5 = $4 \times 3 \times 2 \times 1 = 24$

Possible numbers with unit’s place as 4 and ten’s place 3,2,1 = $3 \times 3 \times 2 \times 1 = 18$

Possible numbers with unit’s place as 3 and ten’s place 2,1 = $2 \times 3 \times 2 \times 1 = 12$

Possible numbers with unit’s place as 3 and ten’s place 1 = $1 \times 3 \times 2 \times 1 = 6$

Total possible values = 24+18+12+6 = 60

Question 20: A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?

a) 0

b) 1

c) 2

d) None of these

Solution:

Let the number ‘n’ belong to the set A.
Hence, the remainder when n is divided by 2 is 1
The remainder when n is divided by 3 is 2
The remainder when n is divided by 4 is 3
The remainder when n is divided by 5 is 4 and
The remainder when n is divided by 6 is 5

So, when (n+1) is divisible by 2,3,4,5 and 6.
Hence, (n+1) is of the form 60k for some natural number k.
And n is of the form 60k-1

Between numbers 0 and 100, only 59 is of the form above and hence the correct answer is 1

Question 21: How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?

a) 0

b) 1

c) 4

d) 3

Solution:

As we know for a number to be divisible by 125, its last three digits should be divisible by 125
So for a five digit number, with digits 2,3,8,7,5 its last three digits should be 875 and 375
Hence only 4 numbers are possible with its three digits as 875 and 375
I.e. 23875, 32875, 28375, 82375
Question 22: What is the digit in the unit’s place of $2^{51}$?

a) 2

b) 8

c) 1

d) 4

Solution:

The last digit of powers of 2 follow a pattern as given below.

The last digit of $2^1$ is 2
The last digit of $2^2$ is 4
The last digit of $2^3$ is 8
The last digit of $2^4$ is 6

The last digit of $2^5$ is 2
The last digit of $2^6$ is 4
The last digit of $2^7$ is 8
The last digit of $2^8$ is 6

Hence, the last digit of $2^{51}$ is 8

Question 23: If $a, b, c,$ and $d$ are integers such that $a+b+c+d=30$ then the minimum possible value of $(a – b)^{2} + (a – c)^{2} + (a – d)^{2}$  is

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 24: If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

a) 1777

b) 1785

c) 1875

d) 1877

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 25: 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

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 26: 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

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

a) -32

b) 16

c) 48

d) 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 28: The smallest integer n for which $4^{n} > 17^{19}$ holds, is closest to

a) 37

b) 35

c) 33

d) 39

Solution:

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

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

Therefore, we can say that n/2 > 19

n > 38

Hence, option D is the correct answer.

Question 29: What is the largest positive integer n such that $\frac{n^2 + 7n + 12}{n^2 – n – 12}$ is also a positive integer?

a) 6

b) 16

c) 8

d) 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 30: 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=105  ==> n=53, m=52

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

n-m=5, n+m=21  ==> 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 31: 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 are $3^0, 3^2, 3^4$ = 3

In $5^4$, 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 32: 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 of 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 33: 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) 58

b) 85

c) 50

d) 95

Solution:

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

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

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

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

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

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

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

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

Adding ab=616 on both sides, we get

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

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

=> a+b=50

Question 34: 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 35: The mean of all 4-digit even natural numbers of the form ‘aabb’,where $a>0$, is

a) 4466

b) 5050

c) 4864

d) 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 36: 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) 49

b) 56

c) 59

d) 46

Solution:

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

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

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

Hence, Option D is the correct answer.

Question 37: 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) 3

b) 1

c) 2

d) 4

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

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

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

Hence m-2n = 9-8 = 1

Question 38: How many integers in the set {100, 101, 102, …, 999} have at least one digit repeated?

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 39: 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 for N?

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 40: How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

a) 42

b) 41

c) 40

d) 43

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, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41

Question 41: How many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ ?

a) 2018

b) 2019

c) 2017

d) 2020

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.

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

Question 42: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

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 43: 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:

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 44: 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

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 place 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.  (4)

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

Since c cannot be greater than 9 considering c to be the maximum value 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 condition is 4195