Quantitative Short Tricks for Problems on Number

Quantitative Aptitude

Numbers

Prime number – which is divisible by only one (1)

Co-prime numbers – which have 1 as common factor or having HCF = 1

Divisibility Rule

For 2:- Check last digit of the number. It should be either ‘0’ or any even number

For 4:- Check last 2 digits. It should be either ‘00’ or divisible by 4.



NOTE

For 2 - 21 - Check last 1 digit

For 4 - 22 - Check last 2 digits



So on...

For 3:- Add all the digits of the number and divide by 3 and check whether it is divisible or not.

For 9:- Add all the digits of the number and divide by 9 and check whether it is divisible or not.

For 5:- Last digit should be either ‘0’ or ‘5’

For 25:- Last digit should be either ‘00’ or divisible by 25

For 6:- The number should be divisible by both ‘2’ and ‘3’

For 11:- If the difference between the sum of its digits at odd places and sum of its digits at even

places is either ‘0’ or divisible by 11.

For 7:- Multiply last digit by 5 and add that number to remaining digits

Example: - check whether 343 is divisible by 7 or not?

Solution: - multiply 3 with 5 and add the resultant (15) with remaining digits (34), we get 49 as

answer which is divisible by 7

For 13:- Multiply last digit by 4

Example: - Check whether 12519 is divisible by 13 or not?

Solution:-



3 4 3

× 15

5

3 4 1 5 4 9

1 2 5 1 9

× 36

4

3 6

1 2 8 7

× 28

4

2 8

1 5 6

× 24

4

2 4 3 9

7 7 5 2

× 24

12

2 4

7 9 9

× 108

12

1 0 8 1 8 7

Which is divisible by 17 So, 7752 is also divisible by 17

Which is divisible by 13 so 12519 So, also divisible by 13

Which is divisible by 7 so, 343 is also divisible by 7

+



Solution: -

6 9 3 5

× 10

2

1 0

7 0 3

× 6

2

6

7 6

× 12

2

1 2

1 9

For 29: - Multiply last digit by 3


Solution: -

1 0 4 4

× 12

3

1 2

1 1 6

× 18

3

1 8

2 9



Cyclicity

Write the last or unit digit

21 - 2 31 - 3 41 - 4 51 - 5

22 - 4 32 - 9 42 - 6 52 - 5

23 - 8 33 - 7 43 - 4 53 - 5

24 - 6 34 - 1 44 - 6

25 - 2 35 - 3

Unit digit of 16 is 6. Similarly unit digit of 32 is 2 and so on...

61 - 6 71 - 7 81 - 8 91 - 9

62 - 6 72 - 9 82 - 4 92 - 1

63 - 6 73 - 3 83 - 2 93 - 9

74 - 1 84 - 6

75 - 7 85 - 8

Clearly, we can see that 1, 5 and 6 are the numbers which always gives same unit digit on

every power.

Small box indicates that after that power the series of digits at unit place repeats.

Example1: - Find the unit digit of the product 784 × 618 × 917 × 463

Solution: - 784 × 618 × 917 × 463

Take the product of unit digits of every no.

= 4 × 8 × 7 × 3

= 32 × 21

= 2 × 1 (again take unit digit only)

= 2

Example 2: - what is the unit digit of 7105

Solution: - The cyclicity of 7 is 4. It means 74 or 7

multiple of 4 gives unit digit 1 (check cyclicity

rule)

we can write 7105 as 7

104 + 1

= 74x(26) ×71

= 1 × 7 (Because 74 or 7

multiple of 4 gives unit digit 1)

So, answer is 7

Example 3: - find the digit at unit place of 365 × 659 × 771

Solution: - They cyclicity of 3 is four, 6 is one and 7 is four

So, 34x(16)+1 × 659 × 74x(17)+3

34x(16)31 × 659 × 74x(17)73

3 × 6 × 3 (Because the unit digit of 34 or 3

multiple of 4 is 1 and that of 73 is 3)

54

So, the answer is 4

Example: - Find the digit at unit place of (4137)754

Solution: - (4137)754

7754 take unit digit only

74x(188)+2

74x(188) × 72

1 × 49 (Because the unit digit of 74 or 7

multiple of 4 is 1)

So, the answer is 9

Some Formulas

(a + b)2 = a2 + b2 + 2ab

(a – b)2 = a2 + b2 – 2ab

(a + b)3 = a3 + b3 + 3ab (a + b)

(a – b)3 = a3 – b3 – 3ab (a – b)

a3 + b3 = (a + b) (a2 – ab +b2)

a3 – b3 = (a – b) (a2 + ab +b2)

a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ac) + 3abc

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ac)

Remainder Therorem

1. an + bn ÷ a + b Divisible only when n = odd

2. an + bn ÷ a – b never divisible

3. an – bn ÷ a + b Divisible only when n = even

4. an – bn ÷ a + b always divisible

Proof

n = Even n = odd

1 (a2 + b2) (a3 + b3)

(a + b) (a + b)

No Formula (a + b)(a2 – ab + b2) (Divisible)

(Not Divisible) (a + b)

Hence, an + bn ÷ a + b Divisible only when n = odd

2 (a2 + b2) (a3 + b3)

(a – b) (a – b)

No Formula (a + b)(a2 – b + b2) (Not Divisible)

(Not Divisible) (a – b)

Hence, an + bn ÷ a – b never divisible

3 (a2 – b2) (a3 – b3)

(a + b) (a + b)

(a + b)(a – b) (a – b)(a2+ ab+b2) (Not Divisible)

(a + b) (a + b)

Hence, an – bn ÷ a + b Divisible only when n = even

4 (a2 – b2) (a3 – b3)

(a – b) (a – b)

(a + b)(a – b) (a – b)(a2+ ab+b2)

(a – b) (a – b)

Hence, an – bn ÷ a + b always divisible

Examples: what is the remainder of

6767 + 67

(67+1)

By adding and subtracting 1

6767 + 1 – 1 + 67

(67+1)

Either (6767 + 1) + (67 – 1) or (6767 – 1) + (67 + 1)

(67+1) (67+1)

(6767 + 167)+ 66 (6767 – 167) + 68

(67+1) (67+1)

6767 + 167 66 6767 – 167 68

(67+1) 68 (67+1) 68

6767 + 167 is divisible as an + bn is Divisible only when n = odd

(67+1) a + b

and

6767 – 167 is not divisible as an - bn is Divisible only when n = even

(67+1) a + b

6767 + 167 66 6767 – 167 68

(67+1) 68 (67+1) 68

So, the remainder is

0 + 66 not divisible

66 is answer

Example: 4915 – 1 is divisible by which of the following

A) 8 B) 14 C) 48 D) 50

Solution:

4915 – 1 4915 – 1 4915 – 1 4915 – 1

8 14 48 50

+ +

+ +

4915 – 1 4915 – 1 4915 – 1 4915 – 1

7 + 1 15 – 1 49 – 1 49 + 1

(72)15 – 1 Not Formula Always Divisible Not Divisible

7 + 1 because n = Odd

730 – 1

7 + 1

Divisible because

n = even

So, 4915 – 1 is divisible by both 8 and 48.

Example: What is remainder when 15 × 16 × 17 is divided by 7?

Solution:

15 × 16 × 17

7

Divide 15, 16 and 17 individually by 7 and multiply the remainder.

Then again divide the resultant by 7

1 × 2 × 3

7

6

7

Now 6 is not divided by 7

So, 6 is the answer

Example: What is remainder when 19 × 20 × 21 is divided by 9?

Solution:

19 × 20 × 21

9

Divide 19, 20 and 21 individually by 9 and multiply the remainder.

Then again divide the resultant by 9

1 × 2 × 3

9

6

9

Now 6 is not divided by 9

So, 6 is the answer (Note: Don’t divide 6 and 9 by 3 to make it lowest fraction)

Example: Out of 12/15 and 15/19 which ratio is bigger?

Solution:

12 15

15 19

By cross multiplication

12 × 19 15 × 17

228 255

Here, 255 is greater than 228 so, 15/19 is greater than 12/15

Example: Out of 3/5 and 2/5 which ratio is bigger?

Solution:

3 2

5 5

By cross multiplication

3 × 5 2 × 5

15 10

Here, 15 is greater than 10 so, 3/5 is greater than 2/5

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Multiplication

1. Multiplication of 2 digit numbers which are close to 100

Case i): When both numbers are less than 100

97

96

97 is 3 below 100 and 96 is 4 below 100

Imagine like below

97 -3 (97 is 3 below 100 thats why –ve sign is placed)

96 -4 (96 is 4 below 100)

Now

a) multiply -3 and -4 and write the resultant in two digit term as 100 contains 2 zero

b) subtract either 4 from 97 or 3 from 96. In both ways resultant will be same.

97 -3

96 -4

93 12 ( –3 × –4 = 12)

( 97 – 4 = 93 or 96 – 3 = 93)

9312 is the answer

Case ii): When both numbers are more than 100

102

104

102 is 2 above 100 and 104 is 4 below 100

Imagine like below

102 +2 (102 is 2 above 100 thats why +ve is sign placed)

104 +4 (104 is 4 below 100)

Now

a) Multiply 2 and 4 and write the resultant in two digit term as 100 contains 2 zero

b) Add either 4 to 102 or 2 to 104. In both ways resultant will be same.

×

×

×

102 +2

104 +4

106 08 ( +2 × +4 = 08)

( 102 + 4 = 106 or 104 + 2 = 106)

10608 is the answer

Case iii): When number of digits of the product of remainder is greater than the power of

10 or two digit in the case of 100.

84

92


Imagine like below


92 -8 (92 is 8 below 100)

Now

a) multiply -16 and -8 and write the resultant in two digit term as 100 contains 2 zero

b) Subtract either 8 from 84 or 16 from 92. In both ways resultant will be same.

84 -16

92 -8

76 128 ( –16 × –8 = 128)

( 84 – 8 = 76 or 92 – 16 = 76)

Carry 1 to 76 to make 128 a two digit term

So, 77 28 is the answer (as 76 + 1 = 77)

×

×

×

Case iii): When one number is greater than 100 & other number is less than 100.

105

93

105 is 5 above 100 and 93 is 7 below 100

Imagine like below

105 +5 (105 is 5 above 100 thats why +ve sign is placed)


Now

a) multiply +5 and – 7 and write the resultant in two digit term as 100 contains 2 zero

b) Either subtract 7 from 105 or add 5 to 93. In both ways resultant will be same.

105 +5

93 -7

98 00

– 35 ( +5 × –7 = –35 thats why we have to minus 35)

( 105 – 7 = 98 or 93 + 5 = 98)

97 65

So, 9765 is the answer

×

×

2. Multiplication of 2 digit numbers which are close to 50

Note: this formula can be used only when both numbers are either even or odd.

47

43


Imagine like below


43 -7 (43 is 7 below 50)

Now

a) multiply -3 and -7 and write the resultant in two digit term

b) Subtract either 7 from 47 or 3 from 43 and divide the resultant by 2 as 50 is equal to

100 divided by 2.

47 -3

43 -7

20 21 ( –3 × –7 = 21)

( 47 – 7 = 40 or 43 – 3 = 40) (40/2 = 20)


58 +8

52 +2

30 16

58 +8

42 –8

25 00

–64

24 36

×

×

47 – 7

2 =

×

58 + 2

2 = 8 × 2

×

58 – 8

2 =

+8 × –8 = –64

Criss cross multiplication:

4 7

4 3

4 2

1 6 0 1

2 0 2 1

Step 1: 4 7

4 × 4 = 16 1 6

4 3

Step 2: 4 7

4 × 3 + 4 × 7 4

4 3 12 + 28 = 40 1 6 0

Write 0 and carry 4

Step 3: 4 7

7 × 3 = 21 4 2

4 3 1 6 0 1

Write 0 and carry 4

Write 1 and carry 2

By adding

4 2

1 6 0 1

2 0 2 1


×

5 4 3 2

3 1 2 4

STEPS

5 × 3 = 15

5 × 1 + 3 × 4 = 17

5 × 2 + 3 × 3 + 4 × 1 = 23

5 × 4 + 3 × 2 + 4 × 2 + 3 × 1 = 37

4 × 4 + 2 × 1 + 3 × 2 = 24

3 × 4 + 2 × 2 = 16

2 × 4 = 8

1 2 3 2 1

1 5 7 3 7 4 6 8

1 6 9 6 9 5 6 8 So, 16969568 is the answer.

STEP – 1

STEP – 7

STEP – 5

STEP – 6

STEP – 2

STEP – 3

STEP – 4

5 4 3 2

3 1 2 4

5 4 3 2

3 1 2 4

×

5 4 3 2

3 1 2 4

5 4 3 2

3 1 2 4

5 4 3 2

3 1 2 4

5 4 3 2

3 1 2 4

5 4 3 2

3 1 2 4

1 5

1

1 5 7

1 2

1 5 7 3

1 2 3

1 5 7 3 7

1 2 3 2

1 5 7 3 7 4

1 2 3 2 1

1 5 7 3 7 4 6

1 2 3 2 1

1 5 7 3 7 4 6 8

Multiplication of a no. with 11 to 19

Multiplication of 21432 with 11

Multiplication with 11

2 1 4 3 2 × 11

Write the first no. from left hand side as it is. And then write (first + second), (second +

third), (third + fourth), (fourth + fifth) and last as it is

2 1 4 3 2 × 11 = 2 3 5 7 5 2

4216 × 11 = 4 6 3 7 6


1 3 3 2 × 1 2 = 1 5 9 8 4


1

2 1 3 2 × 1 3 = 2 7 6 1 6 = 2 7 7 1 6


3 2 1 1

4 3 2 2 × 1 7 = 4 1 3 6 4 = 7 3 4 7 4

2

+

1

1

+

4

4

+

3

3

+

2

2

2×1

+

3

2×3

+

3

2×3

+

2

2×2

3×2

+

1

= 7

3×1

+

3

= 6

3×3

+

2

=11

3×2

=6

7×4

+

3

=31

7×3

+

2

=23

7×2

+

2

=16

7×2

=14

98 + 2

98 - 2

Multiplication of two digit numbers whose ten’s digit is common and their unit’s digits

makes sum as 10:

4 3 × 4 7 = 2 0 2 1

2 4 × 2 6 = 6 2 4

8 5 × 8 5 = 72 2 5

Squaring of a number

(98)2 = 98 is 2 less than nearest “0” term i.e, 100

So,

100

(98)2 = 98 9600 + 4 = 9604

96

(13)2 = 13 is 3 more than nearest “0” term i.e, 10

16

(13)2 = 13 160 + 9 = 169

10 13 – 3

13 + 3

3 × 7

4 × 5

4 × 6

2 × 3

5 × 5

8 × 9

(2)2 = 4

(3)2 = 9


30

(25)2 = 25 600 + 25 = 625

20


134

(132)2 = 132 17420 + 4 = 17424

130

Square of a no. whose unit digit is 5

(15)2 = 2 25

(115)2 = 132 25

(85)2 = 72 25

452 = 4 x 5 / 5 x 5 = 20 / 25 = 2025

552 = 5 x 6 / 5 x 5 = 30 / 25 = 3025

752 = 7 x 8 / 5 x 5 = 56 / 25 = 5625

952 = 9 x 10 / 5 x 5 = 90 / 25 = 9025

25 – 5

25 + 5

(5)2 = 25

132 – 2

132 + 2

(2)2 = 4

(5)2 1 × 2

11 × 12 (5)2

8 × 9 (5)2

Square root

1 = 1 11 = 121

2 = 4 12 = 44

3 = 9 13 = 169

4 = 16 14 = 196

5 = 25 15 = 225

6 = 36 16 = 256

7 = 49 17 = 289

8 = 64 18 = 324

9 = 81 19 = 361

10 = 100 20 = 400

From the above table we can conclude

1, 4, 9, 6, 5, 0 : the no. which ends with these no. will have a perfect square root

Rest of no (2, 3, 7, 8) : the no. which ends with these no. don’t have any perfect

square root

Suppose we have to find out the square root of 24336.

This no. ends with 6, so it will have a perfect square root

243 / 36

Step 1 : separate the last two digits

Step 2 : check the numbers whose square ends with 6

Here in this case 4 and 6 have the squares ends with 6 (i.e, 16 or 36)

Step 3: check the number whose square is nearest smaller than remaining digits

Here in this case (15)2 = 225 is nearest smaller than 243

Write as below

243 36

15 4

or

15 6

so, we comes with two values 154 and 156

the middle term is 155

the square of 155 is 24025

24336 is less than 24025

So the required answer should be (less than 155) 154.

To find square root of any no. upto 2 decimal places

√x = √y ± z

y must be that no. whose square root is known

= √y ± z

2√y

Find √156 find √0.8

√156 = √144 + 12 √0.8 = √1 – 0.2

= √144 + 12 = √1 – 0.2

2√144 2√1

= 12 + 12 = 1 – 0.2

2 × 12 2 × 1

= 12 + 0.5 = 1 – 0.1

= 12.5 = 0.9

Cube of a two digit Number:

Suppose we have to find cube of ab ( a two digit no.)

(ab)3 = a3 a2b ab2 b3

2 a2b 2ab2

a3 3a2b 3ab2 b3

(12)3 = 13 12 ×2 1×22 23

1 2 4 8

4 8

1 7 2 8

(15)3 = 13 12 ×5 1×52 53

1 5 25 125

10 50

1 15 75 125

1

1 7 2

1 5 5 5

3 3 7 5

(45)3 = 43 42 ×5 4×52 53 2 3 1

64 80 100 125 4 0 2

160 200 6 4 0 0 5

64 240 300 125 9 1 1 2 5

Cube root

1 = 1 11 = 1331

2 = 8 12 = 1728

3 = 27 13 = 2197

4 = 64 14 = 2744

5 = 125 15 = 3375

6 = 216 16 = 4096

7 = 344 17 = 4913

8 = 512 18 = 5832

9 = 729 19 = 6859

10 = 1000 20 = 8000

Suppose we have to find out the square root of 32768.

32 / 768

Step 1 : separate the last three digits

Step 2 : check the numbers whose cube ends with 8

Here in this case 2

Step 3: check the number whose cube is nearest smaller than remaining digits

Here in this case (3)3 = 27 is nearest smaller than 32

32 768

3 2

So, the cube root of 32768 is 32.

Arithmetic progression (A.P)

a, a+d, a+2d are said to be in A.P in which first term = a and common difference = d

Nth term = a + (n – 1)d

Sum of n terms = n[2a + (n – 1)d]

2

Sum of n terms = n(a + l) where l is the last term

2

Geometric progression (G.P)

a, ar, ar2, ar3 are said to be in G.P in which first term = a and common ration = r

Nth term = arn-1

Sum of n terms = a(1 – rn) , when r < 1

(1 – r)

a(rn – 1) , when r > 1

(r – 1)

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Quantitative Short Tricks for Problems on Number