Tips, Formulae and Shortcuts for
'Number Systems'
By
CRACKU.IN
Number Systems Tips by CRACKU.IN
▪ Number Systems is the most important topic in the quantitative
section.
▪ It is a very vast topic and a significant number of questions appear
in CAT every year from this section.
▪
Learning simple tricks like divisibility rules, HCF and LCM, prime
number and remainder theorems can help improve the score
drastically.
▪ This document presents best short cuts which makes this topic
easy and helps you perform better.
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Cracku Tip 1 - Number systems
HCF and LCM
▪
HCF * LCM of two numbers = Product of two numbers
▪
The greatest number dividing a, b and c leaving remainders
of x1, x2 and x3 is the HCF of (a-x1), (b-x2) and (c-x3).
▪
The greatest number dividing a, b and c (a<b<c) leaving the same
remainder each time is the HCF of (c-b), (c-a), (b-a).
▪
If a number, N, is divisible by X and Y and HCF(X,Y) = 1. Then, N is
divisible by X*Y
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Cracku Tip 2 - Number systems
Prime and Composite Numbers
▪ Prime numbers are numbers with only two factors, 1 and the
number itself.
▪ Composite numbers are numbers with more than 2 factors.
Examples are 4, 6, 8, 9.
▪ 0 and 1 are neither composite nor prime.
▪There are 25 prime numbers less than 100.
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Cracku Tip 3 - Number systems
Properties of Prime numbers
▪
To check if n is a prime number, list all prime factors less than or
equal to √n. If none of the prime factors can divide n then n is a
prime number.
▪
For any integer a and prime number p, ap−a is always divisible
by p
▪
All prime numbers greater than 2 and 3 can be written in the form
of 6k+1 or 6k-1
▪
If a and b are co-prime then a(b-1) mod b = 1.
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Cracku Tip 4 - Number systems
Theorems on Prime numbers
Fermat's Theorem:
Remainder of a^(p-1) when divided by p is 1, where p is a prime
Wilson's Theorem:
Remainder when (p-1)! is divided by p is (p-1) where p is a prime
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Cracku Tip 5 - Number systems
Theorems on Prime numbers
Remainder Theorem
 If a, b, c are the prime factors of N such that N= ap * bq * cr
Then the number of numbers less than N and co-prime to N is
ϕ(N)= N (1 - 1/a) (1 - 1/b) (1 - 1/c).
This function is known as the Euler's totient function.
Euler's theorem
(Note: If N is prime, the Euler's Theorem becomes the Fermat's Theorem.)
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Cracku Tip 6 - Number systems
▪
Highest power of n in m! is [m/n]+[m/n2]+[m/n3]+…..
Ex: Highest power of 7 in 100! = [100/7] +[100/49] = 16
▪
To find the number of zeroes in n! find the highest power of 5 in n!
▪
If all possible permutations of n distinct digits are added together
the sum = (n-1)! * (sum of n digits) * (11111... n times)
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Cracku Tip 7 - Number systems
▪
If the number can be represented as N = ap ∗ bq∗ cr. . . then
number of factors the is (p+1) * (q+1) * (r+1)
▪
Sum of the factors =
▪
If the number of factors are odd then N is a perfect square.
▪
If there are n factors, then the number of pairs of factors would be
n/2. If N is a perfect square then number of pairs (including the
square root) is (n+1)/2
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Cracku Tip 8 - Number systems
If the number can be expressed as N = 2p ∗ aq ∗ br . . . where
the power of 2 is p and a, b are prime numbers
▪
Then the number of even factors of N = p (1+q) (1+r) . . .
▪
The number of odd factors of N = (1+q) (1+r)…
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Cracku Tip 9 - Number systems
Number of positive integral solutions of the equation x2 - y2 = k is given
by
▪
Total number of factors of k
2
▪
(Total number of factors of k) − 1
2
▪
▪
Total number of factors of
k
4
2
k
4
(If k is odd but not a perfect square)
(If k is even and not a perfect square)
(Total number of factors of ) −1
2
(If k is odd and a perfect square)
(If it is even and a perfect square)
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Cracku Tip 10 - Number systems
▪ Number of digits in ab = [ b logm(a) ] + 1 ; where m is the base
of the number and [.] denotes greatest integer function
▪ Even number which is not a multiple of 4, can never be
expressed as a difference of 2 perfect squares.
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Cracku Tip 11 - Number systems
▪ Sum of first n odd numbers is n2
▪ Sum of first n even numbers is n(n+1)
▪ The product of the factors of N is given by Na/2, where a is
the number of factors
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Cracku Tip 12 - Number systems
▪ The last two digits of a2, (50 - a)2, (50+a)2, (100 - a)2 . . .. . . . . are
same.
▪ If the number is written as 210n
When n is odd, the last 2 digits are 24.
When n is even, the last 2 digits are 76.
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Cracku Tip 13 - Number systems
Divisibility
▪ Divisibility by 2: Last digit divisible by 2
▪ Divisibility by 4: Last two digits divisible by 4
▪ Divisibility by 8: Last three digits divisible by 8
▪ Divisibility by 16: Last four digit divisible by 16
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Cracku Tip 14 - Number systems
Divisibility
▪ Divisibility by 3: Sum of digits divisible by 3
▪ Divisibility by 9: Sum of digits divisible by 9
▪ Divisibility by 27: Sum of blocks of 3 (taken right to left) divisible by 27
▪ Divisibility by 7: Remove the last digit, double it and subtract it from the
truncated original number. Check if number is divisible by 7
▪Divisibility by 11: (sum of odd digits) - (sum of even digits) should be 0
or divisible by 11
Cracku Tip 15 - Number systems
Divisibility properties
▪ For composite divisors, check if the number is divisible by the
factors individually. Hence to check if a number is divisible by 6 it
must be divisible by 2 and 3.
▪ The equation an−bn is always divisible by a-b. If n is even it is
divisible by a+b. If n is odd it is not divisible by a+b.
▪ The equation an+bn , is divisible by a+b if n is odd. If n is even it
is not divisible by a+b.
Cracku Tip 16 - Number systems
• Converting from decimal to base b. Let R1, R2 . . . be the
remainders left after repeatedly dividing the number with b.
Hence, the number in base b is given by ... R2R1.
• Converting from base b to decimal - multiply each digit of the
number with a power of b starting with the rightmost digit and b0.
• A decimal number is divisible by b-1 only if the sum of the digits
of the number when written in base b are divisible by b-1.
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Cracku Tip 17 - Number systems
Cyclicity
▸To find the last digit of an find the cyclicity of a. For Ex. if a=2,
we see that
▸21=2
▸22=4
▸23=8
▸24=16
▸25=32
Hence, the last digit of 2 repeats after every 4th power. Hence
cyclicity of 2 = 4. Hence if we have to find the last digit of an,
The steps are:
1. Find the cyclicity of a, say it is x
2. Find the remainder when n is divided by x, say remainder r
3. Find ar if r>0 and ax when r=0
Cracku Tip 18 - Number systems
▪ (a + b)(a - b) = (a2 - b2)
▪ (a + b)2 = (a2 + b2 + 2ab)
▪ (a - b)2 = (a2 + b2 - 2ab)
▪ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
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Cracku Tip 19 - Number systems
▪
(a3 + b3) = (a + b)(a2 - ab + b2)
▪
(a3 - b3) = (a - b)(a2 + ab + b2)
▪
(a3 + b3 + c3 - 3abc) = (a + b + c)( a2 + b2 + c2 - ab - bc - ac)
▪
When a + b + c = 0, then a3 + b3 + c3 = 3abc.
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Questions
Instructions
For the following questions answer them individually
Question 1
What are the last two digits of
A
21
B
61
C
01
D
41
E
81
72008 ?
Answer: C
 Video Solution
Explanation:
74 = 2401 = 2400+1
So, any multiple of 74 will always end in 01
Since 2008 is a multiple of 4, 72008 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≤x≤6
B
5≤x≤8
C
9 ≤ x ≤ 12
D
11 ≤ x ≤ 14
E
13 ≤ x ≤ 18
Answer: B
 Video Solution
Explanation:
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
A
40
B
37
C
39
100 ≤ n ≤ 200 , are divisible neither by seven nor by nine?
Downloaded from cracku.in
D
38
Answer: C
 Video Solution
Explanation:
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.
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Question 4
The number of positive integers n in the range
A
5
B
7
C
13
D
14
12 ≤ n ≤ 40 such that the product (n -1)*(n - 2)*…*3*2*1 is not divisible by n is
Answer: B
 Video Solution
Explanation:
positive integers n in the range 12 ≤ n ≤ 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
Answer: D
 Video Solution
Explanation:
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?
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A
21
B
25
C
41
D
67
E
73
Answer: C
 Video Solution
Explanation:
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.
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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
Answer: B
 Video Solution
Explanation:
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
Pn denote the product of the digits of n, and Sn denote the sum of the digits of n. The number of integers
between 10 and 1000 for which P n + Sn = n is
A
81
B
16
C
18
D
9
Downloaded from cracku.in
Answer: D
 Video Solution
Explanation:
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
Answer: A
 Video Solution
Explanation:
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 .
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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?
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A
5
B
103
C
6
D
Cannot be determined
Answer: C
 Video Solution
Explanation:
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
Answer: C
 Video Solution
Explanation:
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 4 C3 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 1 a 2 a 1 a 2 a 1 a 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
Answer: C
 Video Solution
Explanation:
Downloaded from cracku.in
Case 1: a 1 = 0
So, D equals 0.0a 2 0a 2 0a 2 ...
So, 100D equals a 2 .0a 2 0a 2 ...
So, 99D equals a 2
Case 2: a 2 = 0
So, D equals 0.a 1 0a 1 0a 1 ...
So, 100D equals a 1 0.a 1 0a 1 ....
So, 99D equals a 1 0
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.
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Question 13
If x2
+ y2 = 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
Answer: B
 Video Solution
Explanation:
(x − y)2 = x2 + y2 − 2xy
0.04 = 0.1 − 2xy => xy = 0.03
So, |xy| = 0.03
(∣x∣ + ∣y∣)2 = x2 + y2 + 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
Answer: C
 Video Solution
Explanation:
The highest power of 5 in 80! = [80/5] + [ 80/52 ] = 16 + 3 = 19
So, the highest power of 5 which divides 80! exactly = 19
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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
Answer: C
 Video Solution
Explanation:
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)
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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
Answer: B
 Video Solution
Explanation:
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.
Hence answer will be B.
Question 17
If n is an integer, how many values of n will give an integral value of
A
2
B
3
C
4
D
None of these
(16n2 +7n+6)
n
?
Answer: D
 Video Solution
Downloaded from cracku.in
Explanation:
6
Expression can be reduced to 16n + 7 + n
Now to make above value an integer n can be 1,2,3,6,-1,-2,-3,-6
Hence answer will be D).
Question 18
n3 is odd. Which of the following statement(s) is/are true?
I. n is odd.
II.n2 is odd.
III.n2 is even.
A
I only
B
II only
C
I and II
D
I and III
Answer: C
 Video Solution
Explanation:
if n3 is odd then n will be odd. let's say it is
2k + 1
then n2 will be = (4k2 + 4k + 1) which will be odd
Hence answer will be C.
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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!
Answer: B
 Video Solution
Explanation:
Possible numbers with unit's place as 5 =
4 × 3 × 2 × 1 = 24
Possible numbers with unit's place as 4 and ten's place 3,2,1 =
3 × 3 × 2 × 1 = 18
Possible numbers with unit's place as 3 and ten's place 2,1 = 2 × 3 × 2 × 1
Possible numbers with unit's place as 3 and ten's place 1 =
= 12
1×3×2×1=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?
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A
0
B
1
C
2
D
None of these
Answer: B
 Video Solution
Explanation:
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
Answer: C
 Video Solution
Explanation:
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
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Question 22
What is the digit in the unit’s place of
A
2
B
8
C
1
D
4
251 ?
Downloaded from cracku.in
Answer: B
 Video Solution
Explanation:
The last digit of powers of 2 follow a pattern as given below.
The last digit of 21 is 2
The last digit of 22 is 4
The last digit of 23 is 8
The last digit of 24 is 6
The last digit of 25 is 2
The last digit of 26 is 4
The last digit of 27 is 8
The last digit of 28 is 6
Hence, the last digit of 251 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
Answer:2
 Video Solution
Explanation:
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
Answer: D
 Video Solution
Explanation:
(x − 1)x(x + 1) = 15600
=> x3 − x = 15600
The nearest cube to 15600 is 15625 = 253
We can verify that x = 25 satisfies the equation above.
Hence the three numbers are 24, 25, 26. Sum of their squares = 1877
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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
Downloaded from cracku.in
Answer:40
 Video Solution
Explanation:
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
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 ≤ 2x ≤ 200 and 2x + 2 is perfectly divisible by either 3 or 4, is
Answer:5
 Video Solution
Explanation:
At x
= 0, 2x = 1 which is in the given range [0.25, 200]
2x + 2 = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution.
At x
= 1, 2x = 2 which is in the given range [0.25, 200]
2x + 2 = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution.
At x
= 2, 2x = 4 which is in the given range [0.25, 200]
2x + 2 = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution.
At x
= 3, 2x = 8 which is in the given range [0.25, 200]
2x + 2 = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution.
At x
= 4, 2x = 16 which is in the given range [0.25, 200]
2x + 2 = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution.
At x
= 5, 2x = 32 which is in the given range [0.25, 200]
2x + 2 = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution.
At x
= 6, 2x = 64 which is in the given range [0.25, 200]
2x + 2 = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution.
At x
= 7, 2x = 128 which is in the given range [0.25, 200]
2x + 2 = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution.
At x
= 8, 2x = 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?
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A
-32
B
16
C
48
D
64
Answer: D
 Video Solution
Explanation:
Let 'a' and 'b' are those two numbers.
⇒ a 2 + b2 = 97
⇒ a 2 + b2 − 2ab = 97 − 2ab
⇒ (a − b)2 = 97 − 2ab
We know that
(a − b)2 ≥ 0
⇒ 97-2ab ≥ 0
⇒ ab ≤ 48.5
Hence, ab

= 64. Therefore, option D is the correct answer.
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Question 28
The smallest integer n for which
A
37
B
35
C
33
D
39
4n > 1719 holds, is closest to
Answer: D
 Video Solution
Explanation:
4n > 1719
⇒ 16n/2 > 1719
Therefore, we can say that n/2 > 19
n > 38
Hence, option D is the correct answer.
Question 29
n2 +7n+12
What is the largest positive integer n such that n2 −n−12 is also a positive integer?
A
6
B
16
Downloaded from cracku.in
C
D
8
12
Answer: D
 Video Solution
Explanation:
n2 +3n+4n+12
n2 −4n+3n−12
n (n+3 )+4 (n+3 )
= n (n−4 )+3 (n−4 )
( n+4 )(n+3 )
= (n−4 )(n+3 )
( n+4 )
= (n−4 )
=
( n−4 )+8
(n−4 )
8
= 1 + (n−4 ) which will be maximum when n-4 =8
n=12
D is the correct answer.
Question 30
How many pairs (m, n) of positive integers satisfy the equation
m2 + 105 = n2 ?
Answer:4
 Video Solution
Explanation:
n2 − m2 = 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.
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Question 31
How many factors of 24
× 35 × 104 are perfect squares which are greater than 1?
Answer:44
 Video Solution
Downloaded from cracku.in
Explanation:
24 × 35 × 104
=24
× 35 × 24 ∗ 54
=28
× 35 × 54
For the factor to be a perfect square, the factor should be even power of the number.
In 28 , the factors which are perfect squares are 20 , 22 , 24 , 26 , 28 = 5
Similarly, in 35 , the factors which are perfect squares are 30 , 32 , 34 = 3
In 54 , the factors which are perfect squares are 50 , 52 , 54 = 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
Answer:7
 Video Solution
Explanation:
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
Answer: C
 Video Solution
Explanation:
Assume the numbers are a and b, then ab=616
We have,
=>
a3 −b 3
(a−b )3 =
157
3
3 (a 3 − b3 ) = 157 (a 3 − b3 + 3ab (b − a ))
=> 154 (a 3
− b3 ) + 3 ∗ 157 ∗ ab (b − a ) = 0
=> 154 (a 3
− b3 ) + 3 ∗ 616 ∗ 157 (b − a ) = 0
=>a 3
(ab=616)
− b3 + (3 × 4 × 157 (b − a )) (154*4=616)
2
2
Downloaded from cracku.in
=> (a
− b) (a 2 + b2 + ab) = 3 × 4 × 157 (a − b)
=> a 2
+ b2 + ab = 3 × 4 × 157
Adding ab=616 on both sides, we get
a 2 + b2 + ab + ab = 3 × 4 × 157 + 616
=> (a
2
+ b) = 3 × 4 × 157 + 616 = 2500
=> a+b=50
CAT Percentile Predictor
Question 34
How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?
Answer:21
 Video Solution
Explanation:
Let the number be 'abc'. Then,
2 < a × b × 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
4466
B
5050
C
4864
D
5544
a > 0, is
Answer: D
 Video Solution
Explanation:
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
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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
Answer: D
 Video Solution
Explanation:
Since c
< 9, we can have the following viable combinations for b × c = 96 (given our objective is to minimize the sum):
48 × 2 ; 32 × 3 ; 24 × 4 ; 16 × 6 ; 12 × 8
× b = 432 into its factors. On close observation, we notice that 18 × 24 and 24 × 4 corresponding to
a × b and b × c respectively together render us with the least value of the sum of a + b + c = 18 + 24 + 4 = 46
Similarly, we can factorize a
Hence, Option D is the correct answer.
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Question 37
Let m and n be natural numbers such that n is even and
A
3
B
1
C
2
D
4
0.2 <
m n n
20 , m , 11
< 0.5. Then m − 2n equals
Answer: B
 Video Solution
Explanation:
0.2 <
n
11
< 0.5
=> 2.2<n<5.5
Since n is an even natural number, the value of n = 4
0.2 <
m
20
Since 0.2
< 0.5 => 4< m<10. Possible values of m = 5,6,7,8,9
<
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?
Answer:252
 Video Solution
Explanation:
Downloaded from cracku.in
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
possible for N?
N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are
Answer:6
 Video Solution
Explanation:
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
Data Interpretation for CAT Questions (download pdf)
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
Answer: B
 Video Solution
Explanation:
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
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Question 41
How many pairs(a, b) of positive integers are there such that
A
2018
B
2019
C
2017
D
2020
a ≤ b and ab = 42017 ?
Answer: A
 Video Solution
Explanation:
ab = 42017 = 24034
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 ≤ b and ab = 42017 = 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?
Answer:315
 Video Solution
Explanation:
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
Logical Reasoning for CAT Questions (download pdf)
Question 43
For all possible integers n satisfying
2.25 ≤ 2 + 2n+2 ≤ 202, then the number of integer values of 3 + 3n+1 is:
Answer:7
 Video Solution
Explanation:
2.25 ≤ 2 + 2n+2 ≤ 202
2.25 − 2 ≤ 2 + 2n+2 − 2 ≤ 202 − 2
0.25 ≤ 2n+2 ≤ 200
Downloaded from cracku.in
log2 0.25 ≤ n + 2 ≤ log2 200
−2 ≤ n + 2 ≤ 7.xx
−4 ≤ n ≤ 7.xx − 2
−4 ≤ n ≤ 5.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 + 3n+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
Answer:4195
 Video Solution
Explanation:
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
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