Since most (20%) of the questions on the CAT paper (the QA part) come from just this chapter, it is important to talk about the details and complexities of the different ideas in an objective and thorough way. Because of this, this part has also gotten very long. So number system is divided into 3 parts. Out of which the first part is Divisibility rules and Basic number syste, concepts.
Students should be well versed with the multiplication and division of numbers without which it will be time taking to understand. Although you will be provided with virtual calculator in the exam, it is always advisable to be very good with calculation of numbers in mind without using pen or paper. It will give an edge in the exam setup and ultimately help you to take better decision in the exam while choosing right questions within the specific time frame.
You shouldn't skip any of it, because each part and idea is just as important as the others.
Lastly, if you want to get ahead in numbers, don't learn by rote. Instead, be curious and open to new ideas. The bottom line is that you should understand and use reasoning because it makes sense.
Checks for Divisibility
We use certain rules to see if a number is divisible by a certain whole number. With the following rules, it's easier to tell if a number is divisible by another number without actually splitting it.
Divisibility by 2
Any whole number whose last digit, called the "unit digit," is even, or whose unit digit can be divided by 2. It means that any number with a last figure of 0, 2, 4, 6, or 8 must be divisible by 2.
Ex:- 242, 454,4234328 etc.
Divisibility by 4
If the number formed by last two digits of the given number is divisible by 4, then the actual number must be divisible by 4. i.e., the last two digits of a number can be 00, 04, 08, 12, 16, 20, 24, 28, 32, ..., 96.
Ex:- 904, 712,14312496 etc.
Divisibility by 8
If the number made by the last three digits of the given number is divisible by 8, then the real number must also be divisible by 8. This means that the last three digits of the divisible number can be 000, 008, 016, 024, 032, 040,..., 096, 104,..., 992.
Ex:- 564008, 657040,5485992 etc.
Divisibility by 16
If the number made by the last four digits of the given number is divisible by 16, then the real number must also be divided by 16. For example, the last four digits of a divisible number can be 0000, 0016, 0032, 0048, 0064, 0080, 0096, 0112, 0128, 0144, 0160,..., 0960, 0976,..., 0992,..., 1600,..., 9984.
Checking if a number can be divided by 32, 64, 128, etc., is as easy as looking at the last 5, 6, 7, etc., digits of the number.
Divisibility by 5
A number is divisible by 5 if and only if the last (unit) figure is either 0 or 5.
Ex:- 05,10,15,20,25 etc.
Divisibility by 3
If the total of a number's digits is divisible by 3, then the number itself is also divided by 3.
Ex:- let 12421412 is the number then 1+2+4+2+1+4+1+2 = 20 so it is not divisible by 3.
Divisibility by 9
If the total of a number's digits is divisible by 9, then the number itself is also divided by 9.
Ex:- 69453 is the number then 6+9+4+5+3= 27. It is divisible by 9 as 27 is the multiple of 9.
Divisibility by 99
Think about any number. Starting from the right side, split the number into pairs like two-digit numbers. If there is a single digit on the left side of the original number, treat it as a single-digit number.
If the total of these numbers is a number that can be divided by 99, then the given number will also be a number that can be divided by 99.
Divisibility by 999
Think about any number. Starting from the right side, split the number into groups like three-digit numbers. If any digits are left ungrouped on the left side of the original number, use the leftover digit(s) as a single or double-digit number. If the total of these numbers is less than 999, then the number given will also be less than 999.
Divisibility by 9999
Think about any number. Starting from the right side, split the number into 4-tuples, which are like four-digit numbers. If there are any digits left on the left side of the original number, use them as a single-, double-, or triple-digit number. If the total of these numbers is a number that can be divided by 9999, then the given number will also be a number that can be divided by 9999.
Divisibility by Some Special Numbers e . g . , 7, 11, 13, 17, 19 etc.
Divisibility by 7
To check the divisibility of a number by 7 we apply the following method. Let the number be 143
First 143 => 14 – 3 x 2 = 8. So 143 is not divisible by 7.
Divisibility by 11
If the difference between the total of the digits at odd places and the sum of the digits at even places is equal to zero or a multiple of 11 (for example, 11, 22, 33, etc.), then the number is divisible by 11.
Ex:- 4289674359
Sum of the digits in the odd places= 4+8+6+4+5 = 27
Sum of the digits in the even places= 2+9+7+3+9= 30
Difference = 30 – 27 = 3
So 4289674359 is not divisible by 11.
Divisibility by 111
Think about any number. Divide the number into triplets starting from the right side as if it were a three-digit number; if any digits are left unpaired on the left side, take the remaining digit(s) as a single or double digit number.
The provided integer will be divisible by 111 if the sum of all these numbers is divisible by 111.
Ex:- Check 759437958473 is divisible by 111
Splitting the number into groups of 3 digits from right to left
759 437 958 473
Adding these numbers as. 759 + 437 + 958 + 473 = 2,627
As 2,627 is not divisible by 111 so 759437958473 is not divisible by 111
Divisibility by 1111
Think about any number. If any digits are left unpaired on the left side of the original number, accept the remaining digit(s) as a single, double, or triple digit number. Split the number into 4-tuples starting from the right side, just as four-digit numbers. The provided integer will also be divisible by 1111 if the sum of these numbers is divisible by 1111.
Divisibility by 13
Let the number is 7235 is divisible by 13
Step 1: 723 5 = 723 + 5 x 4 = 743
Step 2: 74 + 3 x 4= 86
As 86 is not divisible by 13, 7235 is not divisible by 13.
Divisibility by 17
Let the number is 864
Check for Divisibility by 17
Step 1:- 86 4 = 86 – 4 x 5 = 66.
As 66 is not divisible by 17 . So 864 will not be divisible by 17.
Divisibility by 19
Let the number be 84271
Step 1: 8427 1 => 8427 + 1 x 2 = 8429
Step 2: 842 9 => 842 + 9 x 2 = 860
Step 3: 86 0 => 86 + 0 x 2 = 86
As 86 is not divisible by 9. So 84271 will not divisible by 19.
Shortcut rule for the divisibility by 7, 11 and 13
Only when the difference between the number created by the final three digits and the number formed by the remaining digits is divisible by 7, 11, or 13 correspondingly, is a number considered to be divisible by those numbers.
For instance, we must determine whether or not 139125 is divisible by 7.
So, we look at the difference: 139 - 125 = 14
Since the difference is less than 7, it can be divided by 7. So, the given number can also be split into 7 parts.
Divisibility by Composite Numbers e.g., 4, 6, 8, 10, 12, 14, 15 etc.
Only numbers that can be divided by both 2 and 3 can be divided by 6.
So first, we check to see if the number is even or not, and then we see if it can be divided by 3.
Divisibility by 10
The only way a number is divisible by 10 is if it is also divisible by 2 and 5. So, it is easy to see that if a number is divided by 10, it must have one or more zeros at the right end (i.e., the last digits).
Divisibility by 12
A number is only divisible by 12 if it is also divisible by both 4 and 3. So first check if the number can be split by 4 and then by 3.
Divisibility by 15
A number is only divisible by 15 if it is also divisible by both 3 and 5. So divide the number first by 5 and then by 3. So, we can say that if a number is divisible by a composite number, as we saw above, it must also be divisible by all of its factors whose least common multiple is the given divisor.
Squares and cubes
Squaring techniques
It is just knowing how numbers work, so if you know the squares of numbers, you are smarter than people who do not know the squares or who can not figure out the squares quickly in their heads. Do not undermine the role of squares due to your ignorance or impatience.
- Square of the numbers whose unit (i.e., last) digit is zero (0) :
If the last digit of a number is zero, we double the number of zeros on the right and put the square of the numbers that are not zero (or the rest of the digits) on the left of the zeros.
- Square of the numbers whose unit digit is 5 :
- (I) Square of the numbers ending with the digit 1 :
(II) Square of the numbers whose unit digit is 9 :
(III) Square of the numbers whose last digit is 2 or 8 :
(IV) Square of the numbers whose unit digit is 3 or 7 :
(V) Square of the numbers which ends with 4 or 6 :
Properties of Squares
- If the number's unit digit is zero, the unit digit of its square will also be zero, and there will be twice as many zeros in the square as in its root.
- If the number's unit digit is 5, then the unit digit of its square is also 5, and the last two numbers add up to 25.
- If a number's unit digit is 1 or 9, its square will always have a unit digit of 1.
- Whenever the unit digit of a number is 2 or 8, the unit digit of its square is always 4.
- If the unit digit of a number is 3 or 7, its square always has a unit digit of 9.
- If the unit digit of a number is 4 or 6, its square always has a unit digit of 6.
- The square of any number, regardless of the nature of the provided number, is always positive.
- Squaring the numbers with unit digits 0, 1, 5, and 6 yields the same unit digits for each number.
- The unit digits 2, 3, 7, and 8 never appear in the square of a number.
Perfect Square : The square of any natural number is known as perfect square e.g., 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144,....etc.arethesquaresof1,2,3,4,5,6,7,8,9,10,11, 12, ... etc
Square root
The square root of a number is denoted by (n x n), where n2 is derived by multiplying n by itself only once.
There are two methods for finding the square root
- Prime factorization Method
- Division Method
Prime factorization method
In this method, we determine the prime factors first and then link them as shown below.
Ex:- Find the square root of 3600.
The factors of 3600 = (2x2) x (2x2) x (3x3) x (5x5)
Square root of \(\sqrt {3600}\) is = 2 x 2 x 3 x 5 = 60
Division Method
In this method, we begin by forming pairs from the right to the left and then solve as shown below.
Ex:- Square root of 3600
\(\sqrt {3600}\) = 60
Note:
If (11)2 = 121, then (101)2 = 10201
If (21)2 = 441, then
If (201)2 = 40401and so on…
If (11)2 = 121, then (1001)2 = 1002001
If (31)2 = 961, then (3001)2 = 9006001
If (12)2 = 144, then (1002)2 = 1004004 and so on
Cubing Techniques
When a number is multiplied by itself twice, the resulting value is referred to as its cube.
Cube of 4 = 43 = 4 x 4 x 4 = 64
Method to find cube of two digit numbers
Step 1: Jotting down cube of the tens digit
Step 2: Write the the result in the same ratio as the ratio of tens and unit digits in four terms as a G.P.
Step 3: Write the value in the second and third terms below the second and third terms, respectively, is twice the value in the second term.
Step 4: Now you can add up all the numbers in the way shown in the images below:
Cube Roots
The cube root of a number is expressed as \(\sqrt[3]{n}\) or (n)(1/3)
Or n = (n)(1/3) x (n)(1/3) x (n)(1/3)
BASIC Numbers
The following chart shows quickly how the different kinds of numbers are related to each other.
Natural Numbers
The numerals 1, 2, 3, 4, 5,... are referred to as the natural numbers and denoted by N, As you may recall, when a child is very young, he or she begins to count his or her toys, candies, fruits, etc. as 1, 2, 3, 4, 5,..., as he or she has no concept of 0 (zero) and decimal numbers such as 1.5, 2.75, or 3.33 or any other type of numbers. This is likely why the numbers 1, 2, 3, 4, 5, etc. are referred to as natural numbers.
N represents the natural numbers, where N = 1,2,3,K.
All natural numbers are integers that are positive.
Prime Numbers
Except for 1, every natural number that is only divisible by 1 and itself is a prime number, such as 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,... etc.
There are 25 prime integers between 0 and 100. There are 46 prime integers between 100 and 200. The only even prime number and the smallest prime number is two (2). The integer 1 is neither prime nor composite. There are an infinite number of primes.
How to test whether a number is prime or not
To check if a number is a natural number, take its square root and see if it is a natural number. If it isn't, just increase its square root to the next natural number. Then, divide the given number by every prime number that is less than the square root. If the number can be divided by any of these prime numbers, it is not a prime number. Otherwise, it is a prime number.
Co-prime Numbers
Two natural numbers are co-prime (or relatively prime) if they do not share any factors with each other other than 1. The number 1 is the most common factor between two co-prime numbers.
Ex:- (15, 16), (14, 25) etc.
NOTE:- The numbers involved in a pair of co-primes need not be prime; as seen in the preceding examples, they can be composite numbers.
Twin Primes
When the difference between two prime numbers is 2, the numbers are referred to as twin primes.
Ex:- (17, 19), (29, 31) etc.
Composite Numbers
A composite number is any number greater than one that is not a prime number. It indicates that the number is divisible by numbers other than 1 and itself. e.g., 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26...
Every natural number except 1 is either prime or composite.
Whole Numbers
The set of natural numbers that includes '0' is referred to as the set of whole numbers and is denoted by.
W = {0, 1, 2, 3, 4, 5, . . . }
Perfect Number
When the sum of all factors (including 1 but excluding the number itself) of a given number is the same number, we refer to it as a perfect number.
Ex:- 6, 28, 496,etc.
1+2+4+7+14 = 28 Hence 28 is a perfect Number
Triangular Number
Add the preceding number to the nth position in the sequence of triangular numbers, where the first triangular number is 1. Following is the sequence of triangular numbers:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, ....etc.
Previous Year Questions
Ques 1: What is the greatest power of 5 which can divide 80! exactly? (CAT 2016)
- 15
- 16
- 19
- 13
Click Here for the Answer
Ans: (C)
Number of powers of 5 in 80!
=(80/5) +(80/52) =16 + 3 = 19
Ques 2: p, q and r are three non-negative integers such p + q + r = 10. The maximum value of pq+qr+ pr+ pqr. (CAT 2016)
- ≥ 40and< 50
- ≥ 60and< 70
- ≥ 50and< 60
- ≥ 70and< 80
Click Here for the Answer
Ans: (B)
Value of pq + qr + pr + pqr will be maximum, if
Consider p = 3, q = 3, r = 4
pq + qr + pr + pqr = 9+12+12+36=69
Consider, p = 2, q = 3, r = 5
pq + qr + pr + pqr = 6+15+10+30 =61
Thus, we can say that the maximum value of given expression is ≥ 60 and < 70.
Ques 3: N, the set of natural number is partitioned into subsets s1 =(1), s2 =(2,3), s3 =(4,5,6), s4 = (7, 8, 9, 10) and so on. What is the sum of the elements of the subset s50 ? (CAT 2016)
Click Here for the Answer
Ans:
First element of each set = 1, 2, 4, 7, 11, 16, ... . This series is neither an AP nor a GP
the difference between the terms viz. 1, 2, 3, 4, 5, ... is in AP with both first term and common difference as 1.
Hence, to find the 50th term of the original series we have to add the sum of 49th terms of the second series to the first term of the original series.
The sum of first 49th terms = = 1225
Therefore, the 50th term of the original series = (1225 + 1) = 1226
This will be the first element of the set s50, which will have 50 elements.
The last element of s50 will be 1226 + 49 = 1275
So, the sum of the elements in this set
\(\frac{50 (1226+1275)}{2}\) = 62525
Ques 4: If x, y and z are three positive integers such that x > y > z. Which of the following is closest to the product xyz ? (CAT 2016)
- (x−1)yz
- x(y−1)z
- xy(z−1)
- x(y+1)z
Click Here for the Answer
Ans: (A)
Here, yz will be minimum out of yz, xz, xy as x > y > z.
Hence, the correct answer is option (a).
Ques 5: A sequence of 4-digits, when considered as a number in base 10 is four times the number, it represents in base 6. What is the sum of the digits of the sequence? (CAT 2016)
Click Here for the Answer
Ans:
Let the 4-digits sequence be abcd.
In base 6, this represents 216a + 36b + 6c + d and each of a, b, c, d is less than 6.
In base 10, it represents 1000a + 100b + 10c + d.
Given 4 (216a + 36b + 6c + d)=1000a + 100b + 10c+ d
⇒ 136a = 44b + 14c + 3d
By trial a = 1, b = 2, c = 3, d = 2 If a = 2,
LHS = 272
If we consider b = 5, we need 272 − 220 or 52 from 14 c + 3 d (c, d) = (2, 8) but 8 is not a proper digit in base 6.
If a = 3, LHS = 408, while 44b + 14 c + 3 d can at the most be (44 + 14 + 3)5 or 305.
There are no other possible values that satisfy (i).
abcd = 1232 and a + b + c + d = 8
Ques 6: How many pairs of integers (a, b) are possible such that a2 – b2 = 288 (CAT 2016)
Click Here for the Answer
Ans:
(2)1040 /131 = (28 )120 /131 = (256)130 /131
The remainder of a number of the form an , divided by n + 1 (where n + 1 is prime and is relatively prime to a) is always 1.
Hence, the remainder of 21040 divided by 131 is 1.
Ques 7: How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7? (CAT 2020)
- 41
- 42
- 40
- 43
Click Here for the Answer
Ans: (A)
We know that
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).
Therefore, Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) plus Number of Multiples of (5) plus Number of Multiples of (7) - Number of Multiples of (2 and 5) - Number of Multiples of (5 and 7) - Number of Multiples of (7 and 2) + Number of Multiples of (2, 5, and 7)
Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) - Number of Multiples of (10) - Number of Multiples of (35) - Number of Multiples of (14) + Number of Multiples of (70).
Multiples of (2 or 5) or (7) = (60 + 24 + 17 - 12 - 3 - 8 + 1) = 79
The number of integers divisible by at least one of 2, 5, and 7 is 79.
Now, We can determine the number of integers that cannot be divided by 2, 5, or 7 by subtracting 120 from its entirety.
Number of 1, 2,..., 120-digit integers that are not divisible by 2, 5, or 7 = 120 - 79 = 41
Ques 8: How many factors of 24 × 35 × 104 are perfect squares which are greater than 1? (CAT 2019)
Click Here for the Answer
Ans:
24 x 35 x 104
24 x 35 x 24 x 54
28 x 35 x 54
To be a perfect square, a number must contain an even number of factors.
Therefore, if we contemplate the number 28, only the powers of 0, 2, 4, 6, and 8 can result in perfect squares - five possibilities.
If we contemplate the number 35, only the powers of 0, 2, and 4 can lead to perfect squares in three scenarios.
Considering the number 54, only powers of 0.2 and 4 can result in perfect squares in three scenarios.
Therefore, the total number of possibilities = 5 x 3 x 3 = 45
Since we must determine the number of factors greater than 1, we count the number of factors greater than 1.
Required number of methods= 45 - 1 = 44
Ques 9: The number of integers x such that 0.25 < 2x < 200, and 2x + 2 is perfectly divisible by either 3 or 4, is (CAT 2018)
Click Here for the Answer
Ans:
As the intergers are divisible by 3 or 4.
Let us start with value of x as (-2) until 2x is less than 200 and greater than 0.25.
We can see that at x = 0, 1, 2, 4 and 6 the expression 2x + 2 is divisible by either 3 or 4
So the number of integer = 5
Ques 10: 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 Pn + Sn = n (CAT 2014)
Click Here for the Answer
Ans:
Here, 10 < n < 1000
Let n be the two-digit number.
Then, n=10a+b,
Pn =ab, Sn =a+b
∴ ab + a + b = 10 a + b
⇒ ab = 9a
∴ b=9
So, there are 9 two digit numbers, i.e. 19, 29, 39,..., 99. Again, let n be the three-digit number.
n =100a +10b+ c, Pn = a bc, Sn = a + b + c
abc + a + b + c = 100 a + 10 b + c
abc = 99 a + 9 b
bc = 99 + 9 (b/a)
But the maximum value for bc = 81
and RHS is more than 99.
So, no three-digit number is possible. Hence, required number of integers is 9.
Ques 11: The total number of integer pairs (x, y) satisfying the equation x y xy is/are (CAT 2014)
- 0
- 1
- 2
- None of the above
Click Here for the Answer
Ans:
Given,
xy−x−y=0
Adding 1 to both sides of the equation, we get
xy − x − y + 1 = +1
y(x − 1) − 1(x − 1) = 1
( y − 1)(x − 1) = 1 ………. (i)
s xand yare integers, (x−1)and (y−1)are integers.
So, both (x − 1) and ( y − 1) must be 1 or −1 to satisfy Eq. (i),
i.e. x = 2, y = 2 or x = 0, y = 0.
Hence, only two integer pairs satisfy the equation x + y = xy
Ques 12: Suppose n is an integer such that the sum of the digits of n is 2 and 1010<n<1011. The number of different values for n is (CAT 2014)
- 11
- 10
- 9
- 8
Click Here for the Answer
Ans: (A)
We have, 1010 = 10000000000
If anyone of the zeroes is replaced by 1, the value of the result is between 1010 and 1011.
There are 10 zeroes, which can be replaced by 1.2 ×1010.
2 followed by 10 zeroes also lies between 1010 and 1011.
Moreover, the sum of digits of each of the 11 numbers is two. Hence, n is equal to 11.
Ques 13: Let x, y and z be distinct integers, x and y be odd positive and z is even and positive. Which one of the following statements cannot be true? (CAT 2014)
- (x − z)2 y is even
- (x − z) y2 is odd
- (x − z) y is odd
- (x − y)2 z is even
Click Here for the Answer
Ans: (A)
x, y, z > 0; x and y are odd, z is even.
Q [odd − even is odd], [odd − odd is even] and [odd × odd is odd].
Since, (x − z) is odd.
So, (x−z)2 is odd and (x−z)2 y is also odd.
Hence, except option (a) is true.
Ques 14: A three-digit number is eleven times the two-digit number formed by using the hundred’s and the unit’s digit of the three-digit number respectively, in the ten’s and unit’s place of the two-digit number. If the difference between the digit in ten’s place and the digit in hundred’s place is 1, then what is the digit in the unit’s place? (CAT 2014)
- 2
- 3
- 4
- 1
Click Here for the Answer
Ans: (D)
Let the hundred’s, ten’s and unit’s digits be x, y and z, respectively.
Then, (100x + 10 y + z) is the three-digit number
= 11 (10x + z)
= (100x+ z)+ (10x+10z)
∴ y = x + z
Given, y − x = 1
Equating (y-x) from both the equations
∴ z = 1
Note If (x − y) is taken as 1, z = −1 which is inadmissible.
Ques 15: If p, q, r and s are positive real numbers such that p + q + r + s = 2 , then m = ( p + q) (r + s) satisfies the relation (CAT 2013)
- 0 ≤ m ≤ 1
- 1 ≤ m ≤ 2
- 2 ≤ m ≤ 3
- 3 ≤ m ≤ 4
Click Here for the Answer
Ans: (A)
Let p + q = A and r + s = B
p+ q+ r+ s = 2
So, A + B = 2 and AB > 0
AM ≥ GM
\(\frac{p+q}{2}\) ≥ \(\sqrt{ pq}\)
1 ≥ \(\sqrt{ pq}\)
On squaring both sides, we get 1 ≥ pq
pq = m ≤1
∴ 0 ≤ m ≤ 1
How to approach Number system and Divisibility question in CAT exam
- Go through all the divisibility rules and checks. Know it by-heart. More you are thorough with these rules less time consuming in the mocks and CAT exam.
- Remember all the Squares, cubes, square root and cube root till 50.
- Practice as much mocks as possible.
- This chapter’s question are often asked along with concept of Series and sequence so get hold of those concepts as well.
Comments