# Numbers - aptitude test, questions, shortcuts, solved example videos

## Numbers

Important Formulae:

1) Geometric Progression: x, xr, xr3, xr4, --- are said to be in geometric progression. Here, a is first term and r is common ratio.
a) nth term = xr (n -1)
 b) Sum of n terms = x (1 – rn) , here r < 1 (1 – r)
 c) Sum of n terms = x (rn – 1) , here r > 1 (r – 1)

2) Arithmetic Progression: x, x + y, x + 2y, x + 3y are said to be in A.P. Here x is first term and common difference is y.
a) nth term = x + (n – 1) y
 b) Sum of n terms = n [2x + (n – 1)y] 2

 3) 1 + 2 + 3 ± - - - - + n = n(n + 1) 2

 4) (12 + 22 + 32 - - - - + n2) = n(n + 1) (2n + 1) 6

 5) (13 + 23 + 33 - - - - + n3) = n(n + 1) 2 2

Points to Remember:

Difference between Arithmetic Progression and Geometric Progression:
Arithmetic Progression: It is the sequence of numbers in which each term after first is obtained by adding a constant to preceding term. The constant term is called as the common difference.
Geometric Progression: It is a sequence of non-zero numbers. The ratio of any term and its preceding term is always constant.

 Types of Numbers Definition Example Points to remember Natural Numbers Numbers used for counting and ordering 1, 2, 3, 4, 5, ----- natural numbers Whole Numbers All counting numbers along with zero form a set of whole numbers 0, 1, 2, 3, 4 ------ whole numbers Any natural number is a whole number0 is a whole no. which is not a natural no. Integers Counting numbers + negative counting numbers + zero, all are integers -2, -1, 0, 1, 2, ---- integers Positive integers: 0, 1, 2, 3, -------Negative integers: -1, -2, -3, -4, --------- Even Numbers Number divisible by 2 is called as even number 0, 2, 4, 6, 8, ------ even numbers Odd Numbers Number not divisible by 2 is called as even number 1, 3, 5, 7, 9, ------ odd numbers Prime Numbers A number having exactly two factors i.e 1 and itself is called as prime number 2, 3, 5, 7, 11, ----- prime numbers Composite Numbers Natural numbers which are not prime numbers are called as composite numbers 4, 6, 8, 9, 10, ----- composite nos. Co Primes Any two natural numbers x and y are co-prime if their HCF is 1 (4, 5), (7, 9), ---Co-prime numbers

Divisibility of Numbers

1) Number divisible by 2
Units digit – 0, 2, 4, 6, 8
Ex: 42, 66, 98, 1124

2) Number divisible by 3
Sum of digits is divisible by 3
Ex: 267 ---(2 + 6 + 7) = 15
15 is divisible by 3

3) Number divisible by 4
Number formed by the last two digits is divisible by 4
EX: 832
The last two digits is divisible by 4, hence 832 is divisible by 4

4) Number divisible by 5
Units digit is either zero or five
Ex: 50, 20, 55, 65, etc

5) Number divisible by 6
The number is divisible by both 2 and 3
EX: 168
Last digit = 8 ---- (8 is divisible by 2)
Sum of digits = (1 + 6 + 8) = 15 ----- (divisible by 3)
Hence, 168 is divisible by 6

6) Number divisible by 11
If the difference between the sums of the digits at even places and the sum of digits at odd places is either 0 or divisible by 11.
Ex: 4527039
Digits on even places: 4 + 2 + 0 + 9 =15
Digits on odd places: 5 + 7 + 3 = 15
Difference between odd and even = 0
Therefore, number is divisible by 11

7) Number divisible by 12
The number is divisible by both 4 and 3
Ex: 1932
Last two digits divisible by 4
Sum of digits = (1 + 9 + 3 + 2) = 15 ---- (Divisible by 3)
Hence, the number 1932 is divisible by 12

Basic Formulae: (Must Remember)
1) (a - b)2 = (a2 + b2 - 2ab)
2) (a + b)2 = (a2 + b2 + 2ab)
3) (a + b) (a – b) = (a2 – b2 )
4) (a3 + b3) = (a + b) (a2 – ab + b2)
5) (a3 - b3) = (a - b) (a2 – ab + b2)
6) (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
7) (a3 + b3 + c3 – 3abc) = (a + b + c) (a2 + b2 + c2 – ab – bc – ac)

Quick Tips and Tricks:

1) If H.C.F of two numbers is 1, then the numbers are said to be co-prime.
To find a number, say b is divisible by a, find two numbers m and n, such that m*n = a, where m and n are co-prime numbers and if b is divisible by both m and n then it is divisible by a.

2) Sum of the first n odd numbers = n2

3) Sum of first n even numbers = n ( n + 1)

4) Even numbers divisible by 2 can be expressed as 2n, n is an integer other than zero.

5) Odd numbers which are not divisible by 2 can be expressed as 2(n + 1), n is an integer.

6) Dividend = [(Divisor × Quotient)] + Remainder

7) If Dividend = an + bn or an – bn
a) If n is even: an - bn is divisible by (a + b)
b) If n is odd: an + bn is divisible by (a + b)
c) an - bn is always divisible by (a – b)

8) To find the unit digit of number which is in the form ab. (Ex: 7105, 9125)
1) If b is not divisible by 4
Step 1: Divide b by 4, if it is not divisible then find the remainder of b when divided by 4.
Step 2: Units digit = ar, r is the remainder.

2) If b is multiple of 4
Units digit is 6: When even numbers 2, 4, 6, 8 are raised to multiple of 4.
Units digit is 1: When odd numbers 3, 7 and 9 are raised to multiple of 4.

Question Variety

Generally 6 types of numerical are asked from this chapter. Understanding and studying the concepts will help in solving the numerical related to this chapter.

Type 1: Find units digit of a number in the form of ab

Q 1. Find the unit digit of (4137)754

a. 9
b. 7
c. 3
d. 1
View solution

Correct Option:(a)

Hint: Divide b by 4, if it is not divisible then find the remainder of b when divided by 4.
Units digit = ar, r is the remainder
Number is in the form ab i.e (4137) 754
4 × 188 = 752, therefore we get remainder as 2
Units digit = (4137)2 = 17114769
9 is the digit in units place

Q 2. Find the unit digit in the product (365 × 659 × 771)

a. 1
b. 4
c. 5
d. 9
View solution

Correct Option: (b)

Hint:
If b is multiple of 4
Units digit is 6 : When even numbers 2, 4, 6, 8 are raised to multiple of 4.
Units digit is 1 : When odd numbers 3, 7 and 9 are raised to multiple of 4.
Using the hint given, we can easily solve product of large numbers.
[3(4)16 × 3] = (1 × 3) = 3
 = 6
 = [7 (4)17 × 73] = [1 × 3] = 3
Therefore, (3 × 6 × 3) = 54
Required unit digit is 4.

Type 2: Arithmetic and Geometric progression

Q 3. If there are 6 terms in a series, then find the sum of geometric series 2, 6, 18, 54, ----
a. 758
b. 728
c. 754
d. 738
View solution

Correct Option:(b)

 Geometric Progression: Sum of n terms = x (rn – 1) , here r > 1 (r – 1)
Here, x = 2, common ratio (r)= 3 and n = 6
Find the sum of geometric series:
 Sum of n terms = x (rn – 1) , here r > 1 (r – 1)
Substituting the given values,
 Sum of geometric series = 2 (36 – 1) , here r > 1 (3 – 1)
Sum of geometric series = 728

Q 4. Find the number of terms in geometric progression 3, 6, 12, 24, ---- , 384.

a. 10
b. 11
c. 9
d. 8
View solution

Correct Option: (d)

 Here, x = 3, r = 6 = 2 3

We have to find the number of terms
Nth term = 384
nth term = xr (n -1)
384 = 3 x 2 (n -1)
 2(n – 1) = 384 = 128 3
2 (n -1) = 128
27 = 128

Therefore,
2 (n -1) = 27
(n – 1) = 7
n = 7 + 1 = 8

Hence, number of terms = 8

Type 3: Arithmetic Progression

Q 5. If 6 + 12 + 18 + 24 + --- = 1800, then find the number of terms in the series.

a. 21
b. 22
c. 23
d. 24
View solution

Correct Option: (d)

This is an Arithmetic Progression, in which x = 6, y = 6, sum of terms = 1800

 Sum of n terms = n [2x + (n – 1)y] 2
Substituting the given values, we get
 1800 = n [2 × 6 + (n – 1)6] 2

Solving we get,
1800 = 3n (n + 1)
n(n +1) = 600
n2 + n = 600
25 × 24 = 600

Therefore,
n2 + 25n – 24n – 600 = 0
(n + 25) (n – 24) = 0
24 and -25 are the two solutions obtained. Only positive value can be considered. Hence, n = 24
The number of terms in the series = 24

Q 6. 1 + 2 + 3 + ---- 50 = ?

a. 1275
b. 1350
c. 1575
d. 1455
View solution

Correct Option: (a)

This is an Arithmetic Progression, here a = 1, last term = 50 and n = 50

 Sum of n terms = n (a + l) 2

Substituting the given values, we get
 Sum of n terms = 50 (1 + 50) 2
Sum of n terms = 25 × 51 = 1275

Type 4: Divisibility

Q 7. Find which of the following number is divisible by 11?

a. 246542
b. 415624
c. 146532
d. 426513
View solution

Correct Option: (b)

Hint:
Number divisible by 11: If the difference between the sums of the digits at even places and the sum of digits at odd places is either 0 or divisible by 11.
Using the hint, we can easily find the number which is divisible by 11.

Option a) 246542
Sum of digits at even places = (2 + 6 + 4) =12
Sum of digits at odd places = (4 + 5 + 2) = 11
Sum of digits at even places - Sum of digits at odd places = (12 – 11) = 1
Hence, the number is not divisible by 11

Option b) 415624
Sum of digits at even places = (4 + 5 + 2) =11
Sum of digits at odd places = (1 + 6 + 4) = 11
Sum of digits at even places - Sum of digits at odd places = (11 – 11) = 0
Hence, the number is divisible by 11

Option c) 146532
Sum of digits at even places = (1 + 6 + 3) =10
Sum of digits at odd places = (4 + 5 + 2) = 11
Sum of digits at even places - Sum of digits at odd places = (10 – 11) = -1
Hence, the number is not divisible by 11

Therefore, correct answer is option (b) 415624

Q 8. Find the largest 4 digit number which is divisible by 88.

a. 8844
b. 9999
c. 9944
d. 9930
View solution

Correct Option: (c)

We know that the largest 4 digit number is 9999.
Simply divide 9999 by 88.
After dividing 9999 by 88 we get, 55 as remainder.
The number is said to be completely divisible, only if the remainder is zero.
Hence, we can find the required answer by subtracting the remainder obtained from the 4 digit number.
Therefore, required number = 9999 – 55 = 9944

Type 5: Find result of different operations (additions, subtractions, multiplications, divisions etc) on given integers.

Q 9. Find the solution of (935421 × 625) = ?

a. 584638125
b. 524896335
c. 542879412
d. 582365890
View solution

Correct Option: (a)

In this type of questions, try to simplify the numbers, to find the answer easily.
We are asked to multiply (935421 × 625)
625 is divisible by 5 and 54= 625
This 54 can be written as (10 /2) 4

 Hence, 935421 × 104 = 9354210000 = 584638125 24 16

 Q 10. Find the solution of (450 + 280)2 – (450 – 280) 2 (450 × 280)

a. 21
b. 10
c. 7
d. 4
View solution

Correct Option: (d)

Remember the basic formula:
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
If we substitute these formulae, we get

 (a + b)2 – (a – b)2 = (a2 + 2ab + b2 – a2 + 2ab – b2) = 4ab = 4 (a × b) (a × b) ab

Hence, the answer can be determined very easily in this way, instead of multiplying and dividing the terms.

Type 6: Number a when divided by b gives remainder r, what will be the remainder when a is divided by c.

Q 11. The remainder is 29, when a number is divided 56. If the same number is divided by 8, then what is the remainder?
a. 3
b. 4
c. 7
d. 5
View solution

Correct Option : (d)

We know that,
Dividend = [(Divisor × Quotient)] + Remainder
It is given that, the remainder is 29, when a number (dividend) is divided 56(divisor).
Dividend and quotient are unknown, hence assume dividend as X and quotient as Y.
Therefore,
X = 56 × Y + 29
56 is completely divisible by 8, but 29 is not completely divisible and we get remainder as 5, which is the required answer.
OR
X = 56 × Y + 29
= (8 × 7Y) + (8 × 3) + 5
5 is the required remainder.

Q 12. The remainder is 3, when a number is divided 5. If the square of this number is divided by 5, then what is the remainder?

a. 5
b. 4
c. 7
d. 1
View solution

Correct Option : (b)

We know that,
Dividend = [(Divisor × Quotient)] + Remainder
It is given that, the remainder is 3, when a number (dividend) is divided 5(divisor).
Dividend and quotient are unknown, hence assume dividend as X and quotient as Y.
X = (5Y) + 3
The square of this number is divided by 5, therefore
X2 = (5Y + 3)2
Simplifying we get,
X2 = (25Y2 + 30Y + 9)
On dividing this equation, we get
5(5Y2 + 6Y + 1) + 4
The remainder is 4, because 9 is not exactly divisible by 5 and we get 4 as remainder.

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