fdaytalk · fdaytalk.com author: j maha laxmaiah mail id: [email protected] 11 × 13 = 143 12...

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Author: J Maha Laxmaiah Mail id: [email protected]

Natural numbers

Counting numbers 1, 2, 3, 4, 5 ……….…… are known as natural numbers

Whole numbers

If we include zero among the natural numbers, then the numbers 0, 1, 2, 3, 4, 5

……………….. are called whole numbers

Rational Numbers

The numbers of the form 𝒙

𝒚 where X & Y are integers and Y ≠ 0 are known as

rational numbers

Example: 𝟐

𝟑,

𝟓

𝟕,

−𝟒

𝟗 etc.

Irrational numbers

Those numbers which when expressed in decimal form are neither terminating nor

repeating decimals, are known as irrational numbers.

Examples: √𝟐, √𝟑, √𝟓, 𝝅 etc.

Composite numbers

Natural numbers greater 1 which are not prime numbers are composite numbers

Example: 4, 6, 9, 15 etc.

Co – prime numbers

Two numbers which have only 1 as the common factors are called co- prime

numbers or relatively prime to each other.

Example: (3, 7), (8, 9), (36, 25) etc.

Learn the Tables from 11 to 20 (11 × 1 to 11 × 20 format)

11 × 1 = 11 12 × 1 = 12 13 × 1 = 13

11 × 2 = 22 12 × 2 = 24 13 × 2 = 26

11 × 3 = 33 12 × 3 = 36 13 × 3 = 39

11 × 4 = 44 12 × 4 = 48 13 × 4 = 52

11 × 5 = 55 12 × 5 = 60 13 × 5 = 65

11 × 6 = 66 12 × 6 = 72 13 × 6 = 78

11 × 7 = 77 12 × 7 = 84 13 × 7 = 91

11 × 8 = 88 12 × 8 = 96 13 × 8 = 104

11 × 9 = 99 12 × 9 = 108 13 × 9 = 117

11 × 10 = 110 12 × 10 = 120 13 × 10 = 130

11 × 11 = 121 12 × 11 = 132 13 × 11 = 143

11 × 12 = 132 12 × 12 = 144 13 × 12 = 156

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11 × 13 = 143 12 × 13 = 156 13 × 13 = 169

11 × 14 = 154 12 × 14 = 168 13 × 14 = 182

11 × 15 = 165 12 × 15 = 180 13 × 15 = 195

11 × 16 = 176 12 × 16 = 192 13 × 16 = 208

11 × 17 = 187 12 × 17 = 204 13 × 17 = 221

11 × 18 = 198 12 × 18 = 216 13 × 18 = 234

11 × 19 = 209 12 × 19 = 228 13 × 19 = 247

11 × 20 = 220 12 × 20 = 240 13 × 20 = 260

14 × 1 = 14 15 × 1 = 15 16 × 1 = 16

14 × 2 = 28 15 × 2 = 30 16 × 2 = 32

14 × 3 = 42 15 × 3 = 45 16 × 3 = 48

14 × 4 = 56 15 × 4 = 60 16 × 4 = 64

14 × 5 = 70 15 × 5 = 75 16 × 5 = 80

14 × 6 = 84 15 × 6 = 90 16 × 6 = 96

14 × 7 = 98 15 × 7 = 105 16 × 7 = 112

14 × 8 = 112 15 × 8 = 120 16 × 8 = 128

14 × 9 = 126 15 × 9 = 135 16 × 9 = 144

14 × 10 = 140 15 × 10 = 150 16 × 10 = 160

14 × 11 = 154 15 × 11 = 165 16 × 11 = 176

14 × 12 = 168 15 × 12 = 180 16 × 12 = 192

14 × 13 = 182 15 × 13 = 195 16 × 13 = 208

14 × 14 = 196 15 × 14 = 210 16 × 14 = 224

14 × 15 = 210 15 × 15 = 225 16 × 15 = 240

14 × 16 = 224 15 × 16 = 240 16 × 16 = 256

14 × 17 = 238 15 × 17 = 255 16 × 17 = 272

14 × 18 = 252 15 × 18 = 270 16 × 18 = 288

14 × 19 = 266 15 × 19 = 285 16 × 19 = 304

14 × 20 = 280 15 × 20 = 300 16 × 20 = 320

17 × 1 = 17 18 × 1 = 18 19 × 1 = 19

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17 × 2 = 34 18 × 2 = 36 19 × 2 = 38

17 × 3 = 51 18 × 3 = 54 19 × 3 = 57

;17 × 4 = 68 18 × 4 = 72 19 × 4 = 76

17 × 5 = 85 18 × 5 = 90 19 × 5 = 95

17 × 6 = 102 18 × 6 = 108 19 × 6 = 114

17 × 7 = 119 18 × 7 = 126 19 × 7 = 133

17 × 8 = 136 18 × 8 = 144 19 × 8 = 152

17 × 9 = 153 18 × 9 = 162 19 × 9 = 171

17 × 10 = 170 18 × 10 = 180 19 × 10 = 190

17 × 11 = 187 18 × 11 = 198 19 × 11 = 209

17 × 12 = 204 18 × 12 = 216 19 × 12 = 228

17 × 13 = 221 18 × 13 = 234 19 × 13 = 247

17 × 14 = 238 18 × 14 = 252 19 × 14 = 266

17 × 15 = 255 18 × 15 = 270 19 × 15 = 285

17 × 16 = 272 18 × 16 = 288 19 × 16 = 304

17 × 17 = 289 18 × 17 = 306 19 × 17 = 323

17 × 18 = 306 18 × 18 = 324 19 × 18 = 342

17 × 19 = 323 18 × 19 = 342 19 × 19 = 361

17 × 20 = 340 18 × 20 = 360 19 × 20 = 380

Learn the Square of a numbers from 1 to 99

012 = 01 512 = 2601, 112 = 121 612 = 3721

022 = 04 522 = 2704, 122 = 144 622 = 3844

032 = 09 532 = 2809, 132 = 169 632 = 3969

042 = 16 542 = 2916, 142 = 196 642 = 4096

052 = 25 552 = 3025, 152 = 225 652 = 4225

062 = 36 562 = 3136, 162 = 256 662 = 4356

072 = 49 572 = 3249, 172 = 289 672 = 4489

082 = 64 582 = 3364, 182 = 324 682 = 4624

092 = 81 592 = 3481, 192 = 361 692 = 4761

212 = 441 712 = 5041, 312 = 961 812 = 6561

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222 = 484 722 = 5184, 322 = 1024 822 = 6724

232 = 529 732 = 5329, 332 = 1089 832 = 6889

242 = 576 742 = 5476, 342 = 1156 842 = 7056

252 = 625 752 = 5625, 352 = 1225 852 = 7225

262 = 676 762 = 5776, 362 = 1296 862 = 7396

272 = 729 772 = 5929, 372 = 1369 872 = 7569

282 = 784 782 = 6084, 382 = 1444 882 = 7744

292 = 841 792 = 6241, 392 = 1521 892 = 7921

412 = 1681 912 = 8281 462 = 2116 962 = 9216

422 = 1764 922 = 8464 472 = 2209 972 = 9409

432 = 1849 932 = 8649 482 = 2304 982 = 9604

442 = 1936 942 = 8836 492 = 2401 992 = 9801

452 = 2025 952 = 9025

Learn the Cube of a numbers from 1 to 30

13 = 1 113 = 1331 213 = 9261

23 = 8 123 = 1728 223 = 10648

33 = 27 133 = 2197 233 = 12167

43 = 64 143 = 2744 243 = 13824

53 = 125 153 = 3375 253 = 15625

63 = 216 163 = 4096 263 = 17576

73 = 343 173 = 4913 273 = 19683

83 = 512 183 = 5832 283 = 21952

93 = 729 193 = 6859 293 = 24389

103 = 1000 203 = 8000 303 = 27000

Find out square when unit digit is 5

252 = ?

252 = 625

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= (2×3) (52)

= 6 25

= 625 (answer)

352 = ?

352 = 1225

= (3×4) (52)

= 12 25

= 1225 (answer)

452 = ?

452 = 2025

= (4×5) (52)

= 20 25

= 2025 (answer)

852 = ?

852 = 7225

= (8×9) (52)

= 72 25

= 7225 (answer)

952 = ?

952 = 9025

= (9×10) (52)

= 90 25

= 9025 (answer)

Find out a square root of a number

√8464 = ?

√8464 = 92

8464 ≅ 8100

> 902

= 922

Therefore, √8464 = 92

√𝟗𝟒𝟎𝟗 = ?

√9409 = 97

9409 ≅ 9025 (we know that, 952 = 9025)

> 952

= 97

Therefore, √9409 = 97 (answer)

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Find out a square of a number

(106)2 = ?

(106)2 = 11236

Step 1: (06)2 = 36

Step 2: 1× 2× 06 = 12

Step 3: 12 = 1

Final answer = 1 12 36

= 11236

Therefore, (106)2 = 11236 (answer)

(113)2 = ?

(113)2 = 12769

Step 1: (13)2 = 69 (here, 1 is parity or excess)

Step 2: 1× 2× 13 = 26

= 26 + 1 (parity is added here)

= 27

Step 3: 12 = 1


= 12769


(209)2 = ?

(209)2 = 43681

Step 1: (09)2 = 81

Step 2: 2 × 2 × 09 = 36

Step 3: 22 = 4


= 43681


(216)2 = ?

(216)2 = 46656

Step 1: (16)2 = 56 (here 2 is parity or excess)

Step 2: 2 × 2 × 16 = 64

= 64 + 2 (parity 2 is added here)

= 66

Step 3: ( 22 ) = 4


= 46656


To find out a cube root of a number

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√17283

= ?

√17283

= 12

Step 1: in 1728, 8 replaces by 2 (83 = 512)

Therefore, unit place digit is = 2

Step 2: in 1728, ignore 728 (last 3 – digits)

We now have 1

Here, 1 ≅ 13 (1 ≠ 23 )

Therefore, tenth place digit is = 1

Therefore,

√17283

= 12 (answer)

√𝟏𝟗𝟔𝟖𝟑𝟑

= ?

√196833

= 27

Step 1: in 19683, 3 replaces by 7 (33 = 27)

Therefore, unit place digit is = 7

Step 2: in 19683, ignore 683 (last 3 – digits)

We now have 19

Here, 19 ≅ 23 (19 ≠ 33 )

Therefore, tenth place digit is = 2

Therefore,

√196833

= 27 (answer)

TO FIND OUT A CUBE OF A NUMBER

TYPE 1: Number starts with ‘1’ (from left)

Example: 12, 13, 14, 15, ………………….

(12)3 = ?

Step 1: write given number as it as with some space

1 2

Step 2: Square and Cube the unit digit (2) of a given

number (12) and write right side to 1 2

1 2 22 23

= 1 2 4 8

Step 3: Double the middle numbers (2 & 4 only) and

add to them in the same position

1 2×2 4×2 8

= 1 4 8 8

1 2 4 8

add here 4 8

-------------------------------------------

1 7 (1)2 8 [here, parity or excess ‘1’

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added to next left column]

1 7 2 8

So, (12)3 = 1728 answer)

(13)3 = ?


1 3



1 3 32 33

= 1 3 9 27



1 3×2 9×2 27

= 1 6 18 27

1 3 9 27

add here 6 18

-------------------------------------------

2 (1)1 (2)9 (2)7 [here, parity or excess 2, 2

& 1 shown in the brackets added to next left columns respectively]

2 1 9 7

So, (13)3 = 2197 answer)

(15)3 = ?


1 5



1 5 52 53

= 1 5 25 125


add them in the same position

1 5×2 25×2 125

= 1 10 50 125

1 5 25 125

add here 10 50

---------------------------------------------------

3 (2)3 (8)7 (12)5 [here, parity or excess 12,

8 & 2 shown in the bracket added to next left columns respectively]

3 3 7 5

So, (15)3 = 3375 answer)

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TYPE 2: Number ends with ‘1’ (from right)

Example: 21, 31, 41, 51 …………………….

(21)3 = ?


2 1

Step 2: Square and Cube the 10th place digit (2) of a given number (21) and

write left side to 2 1

23 22 2 1

= 8 4 2 1



8 4×2 2×2 1

= 8 8 4 1

8 4 2 1

add here 8 4

-------------------------------------------

9 (1)2 6 1 [here, parity or excess ‘1’

shown in the bracket added to next left columns respectively]

9 2 6 1

So, (21)3 = 9261 answer)

(41)3 = ?


4 1

Step 2: Square and Cube the 10th place digit (4) of a given number (41) and

write left side to 4 1

43 42 4 1

= 64 16 4 1


add them in the same position

64 16×2 4×2 1

= 64 32 8 1

64 16 4 1

add here 32 8

-------------------------------------------

68 (4)9 (1)2 1 [here, parity or excess 1 & 4

shown in the brackets added to next left columns respectively]

68 9 2 1

So, (41)3 = 68921 answer)

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TYPE 3: If both the numbers same

Example: 22, 33, 44, 55 …………………..

(22)3 = ?

Step 1: Here, common number is 2

So, write cube of 2 (i. e 23 = 8) in four places with spaces

8 8 8 8

Step 2: Double the middle numbers (8 & 8) add them to the same position

8 8×2 8×2 8

= 16 16

8 8 8 8

add here 16 16

---------------------------------------------

10 (2)6 (2)4 8 [here, parity or excess 2 & 2

shown in the brackets added to next left columns respectively]

10 6 4 8

So, (22)3 = 10648

(33)3 = ?

Step 1: Here, common number is 3

So, write cube of 3 (i. e 33 = 27) in four places with some spaces

27 27 27 27


27 27×2 27×2 27

= 54 54

27 27 27 27

add here 54 54

---------------------------------------------

35 (8)9 (8)3 (2)7 [here, parity or excess 2, 8

& 8 shown in the brackets added to next left columns respectively

35 9 3 7

So, (33)3 = 35937

TYPE 4: If both the numbers are different

Example: 23, 42, 52, 47, 89, ………………………

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(23)3 = ?

Step 1: write the cube of both the numbers 2 & 3 (i. e 23 = 8 & 33 = 27) with the

some space

8 27

Step 2: in the middle 8 & 27, write like 22 × 3 & 32 × 2

8 22 × 3 32 × 2 27

= 8 12 18 27


8 12×2 18×2 27

= 24 36

8 12 18 27

add here 24 36

-------------------------------------------------

12 (4)1 (5)6 (2)7 [Here, parity or excess 2, 5

& 4 shown in the brackets added to next left columns respectively]

12 1 6 7

So, (23)3 = 12167

(35)3 = ?

Step 1: write the cube of both the numbers 3 & 5 (i. e 33 = 27 & 53 = 125) with

the some space

27 125

Step 2: in the middle 27 & 125, write like 32 × 5 & 52 × 3

27 32 × 5 52 × 3 125

= 27 45 75 125


27 45×2 75×2 125

= 90 150

27 45 75 125

add here 90 150

-------------------------------------------------

42 (15)8 (23)7 (12)5 [Here, parity or excess 12,

23 & 15 shown in the brackets added to next left columns respectively]

42 8 7 5

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So, (35)3 = 42875

WHEN SUM OF THE UNIT DIGIT IS ‘10’

1) 56 × 54 = ?

Step 1: Multiplication of unit digits i.e 6 × 4

Step 2: Multiplication of 10’s digit, here 5 and its next number i.e 5 × 6

(5 × 6) (6 × 4)

30 24

= 3024 (answer)

2) 72 × 78 = ?

Step 1: Multiplication of unit digits i. e 2 × 8

Step 2: Multiplication of 10’s digit, here 7 and its next number i. e 7 × 8

(7 × 8) (2 × 8)

56 16

= 5616 (answer)

3) 113 × 117 = ?

Step 1: Multiplication of unit digits i. e 3 × 7

Step 2: Multiplication of 10’s digits, here 11 and its next number i. e 11 × 12

(11 × 12) (3 × 7)

132 21

= 13221 (answer)

ANY TWO DIGIT MULTIPLICATION

1) 42 × 36 = ?

42 × 36 = ____ ____ ____

Step 1: Multiplication of unit digits ____ ____ 2 × 6

____ ____ 12

Step 2: Sum of multiplication of Extreme numbers and middle numbers i. e (4 × 6)

+ (2 × 3) = 30

____ 30 12

Step 3: Multiplication of 10’s digits, i. e (4 × 3) = 12

12 30 12

In the above, from right, 1 is treated as excess or parity and it has to added to next

number 30

Now, 12 31 2

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From the above, 3 is treated as excess or parity and it has to added to next number

12

Now, 15 3 2

= 1532 (answer)

2) 96 × 73 = ?

96 × 73 = ____ ____ ____

Step 1: Multiplication of unit digits ____ ____ 6 × 3

____ ____ 18

Step 2: Sum of multiplication of Extreme numbers and middle numbers i. e (9 × 3)

+ (6 × 7) = 69

____ 69 18

Step 3: Multiplication of 10’s digits, i. e (9 × 7) = 63

63 69 18

In the above, from right, 1 is treated as excess or parity and it has to added to next

number 69

Now, 63 70 2

From the above, 7 is treated as excess or parity and it has to added to next number

63

Now, 70 0 8

= 7008 (answer)

HOW TO LEARN & WRITE TABLES FROM 11 TO 99:

Table of 12

12

To write a table of 12, just write the tables of 1 & 2 in two separate columns

1 2 = 12

2 4 = 24

3 6 = 36

4 8 = 48

5 10 (here, 10’s place digit is added to 5)

(5 + 1) 0 = 60

6 12 = 72 [6 + 1 = 7]

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7 14 = 84 [7 + 1 = 8]

8 16 = 96 [8 + 1 = 9]

9 18 = 108 [9 + 1 = 10]

10 20 = 120 [10 + 2 = 12]

Table of 26

26


2 6 = 26

4 12 = 52

6 18 = 78

8 24 (here, 10’s place digit is added to 8

(8 + 2) 4 = 104


(10 + 3) 0 = 130

12 36 = 156 [12 + 3 = 15]

14 42 = 182 [14 + 4 = 18]

16 48 = 208 [16 + 4 = 20]

18 54 = 234 [18 + 5 = 23]

20 60 = 260 [20 + 6 = 26]

Table of 94

94


9 4 = 94

18 8 = 188


(27 + 1) 2 = 282

36 16 = 376 [36 + 1 = 37]

45 20 = 470 [45 + 2 = 47]

54 24 = 564 [54 + 2 = 56]

63 28 = 658 [63 + 2 = 65]

72 32 = 752 [72 + 3 = 75]

81 36 = 846 [81 + 3 = 84]

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90 40 = 940 [90 + 4 = 94]

DIFFERENCE IS ‘10’ AND UNIT DIGIT OR ENDS WITH 5:

When the difference of given numbers is 10 and unit digit is 5, the number 75 is

come right side as common

1) 35 × 45 = ?

Step 1: The number 75 is come right side ______ 75

Step 2: The left side number is, square the larger number and substract the ‘1’, i. e

(42 – 1) = 15

= 15 75

= 1575 (answer)

2) 75 × 85 = ?

Step 1: The number 75 is comes right side _____ 75


(82 – 1) = 63

= 63 75

= 6375 (answer)

3) 135 × 145 = ?

Step 1: The number 75 is comes right side _____ 75


(142 – 1) = 195

= 195 75

= 19575 (answer)

SAME NUMBERS AND ENDS WITH ‘5’:

In this case, the number 25 is comes right side as common

1) 65 × 65 = ?

Step 1: The number 25 comes right side _____ 25

Step 2: Multiplication of 10’s place digit and its next number i. e (6 × 7) = 42

= 42 25

= 4225 (answer)

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2) 125 × 125 = ?

Step 1: The number 25 comes right side _____ 25

Step 2: Multiplication of 10’s place digit and its next number i. e (12 × 13) = 156

= 156 25

= 15625 (answer)

ANY NUMBER AND ENDS WITH ‘5’

In this case, when sum of 10’s digits is even number ‘25’ is taken as unit digit

(write right side) in the final answer

When sum of 10’s digits is odd number, ‘75’ taken as unit digit (write right side) in

the final answer

1) 45 × 65 = ?

Step 1: Here, sum of 10’s digit is even (6 + 4 = 10), So, 25 write the right side

_____ 25

Step 2: Multiplication of 10’s digit i. e 4 × 6 = 24

And half the sum of the 10’s digit i. e (4+6

2) = 5

Step 3: add the 24 and 5 => 24 + 5 = 29

= 29 25

= 2925 (answer)

2) 95 × 75 = ?

Step 1: Here, sum of 10’s digit is even (9 + 7 = 16), So, 25 write the right side

_____ 25



2) = 8

Step 3: add the 63 and 8 => 63 + 8 = 71

= 71 25

= 7125 (answer)

3) 35 × 85 = ?

Step 1: Here, sum of 10’s digit is odd (3 + 8 = 11), So, 75 write the right side

_____ 75



2) = 5

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Step 3: add the 24 and 5 => 24 + 5 = 29

= 29 75

= 2975 (answer)

MULTIPLY OF TWO NUMBERS DIFFERING 2, 4, 6, 8, 10, ………

1) [Square of middle number of given numbers] – [1]2

11 × 13 = ?

Here, the difference between 11 & 13 is = 2 and the middle number is 12

Therefore, 122 – 12 => 144 – 1 = 143 (answer)

15 × 17 = ?



24 × 26 = ?




11 × 15 = ?



17 × 21 = ?



60 × 64 = ?




11 × 17 = ?


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13 × 19 = ?



4) [Square of a middle number of given numbers] – [4]2

11 × 19 = ?



14 × 22 = ?



NUMBERS MULTIPLY BY 5, 25, 50, 125, 625:

MULTIPLY BY 5

1) 728 × 5 = ?

728 × 5 = 728 × 5 × 2

2

= 728

2∗ 10

= 3640 (answer)

2) 176 × 5 = ?

176 × 5 = 176 × 5 × 2

2

= 176

2∗ 10

= 880 (answer)

MULTIPLY BY 25

1) 728 × 25 = ?

728 × 25 = 728 × 25 × 4

4

= 728

4∗ 100

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= 18200 (answer)

2) 176 × 25 = ?

176 × 25 = 176 × 25 × 4

4

= 176

4∗ 100

= 4400 (answer)

MULTIPLY BY 50

1) 728 × 50 ?

728 × 50 = 728 × 50 × 2

2

= 728

2∗ 100

= 36400 (answer)

2) 176 × 50 = ?

176 × 50 = 176 × 50 × 2

2

= 728

2∗ 100

= 36400 (answer)

MULTIPLY BY 125

1) 728 × 125 = ?

728 × 125 = 728 × 125 × 8

8

= 728

8∗ 1000

= 91000 (answer)

2) 176 × 125 = ?

176 × 125 = 176 × 125 × 8

8

= 176

8∗ 1000

= 22000 (answer)

NUMBERS DIVISIBLE BY 5, 25, 50, 125, 625

DIVISIBLE BY 5

1) 164

5= ?

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164

5=

164

5∗

2

2

= 164

10∗ 2

= 328

10

= 32. 8 (answer)

2) 624

5= ?

624

5=

624

5∗

2

2

= 624

10∗ 2

= 1248

10

= 124. 8 (answer)

DIVISIBLE BY 25

1) 164

25= ?

164

25=

164

25∗

4

4

= 164

100∗ 4

= 656

100

= 6. 56 (answer)

2) 624

25= ?

624

25=

624

25∗

4

4

= 624

100∗ 4

= 2496

100

= 24. 96 (answer)

DIVISIBILE BY 50

1) 164

50= ?

164

50=

164

50∗

2

2

= 164

100∗ 2

= 328

100

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= 3. 28 (answer)

2) 624

50= ?

624

50=

624

50∗

2

2

= 624

100∗ 2

= 1248

100

= 12. 48 (answer)

DIVISIBILITY BY 125

1) 164

125= ?

164

125=

164

125∗

8

8

= 164

1000∗ 8

= 1312

1000

= 1. 312 (answer)

2) 624

125= ?

624

125=

624

125∗

8

8

= 624

1000∗ 8

= 4992

1000

= 4. 992 (answer)

332 = 1 0 8 9

= 1089

3332 = 11 0 88 9

= 110889

33332 = 111 0 888 9

= 11108889

333332 = 1111 0 8888 9

= 1111088889

992 = 9 8 0 1

= 9801

9992 = 99 8 00 1

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= 998001

99992 = 999 8 000 1

= 99980001

999992 = 9999 8 0000 1

= 9999800001

112 = 1 2 1

= 121

1112 = 12 3 21

= 12321

11112 = 123 4 321

= 1234321

111112 = 1234 5 4321

= 123454321

111, 222, 333, 444, 555 …………….. are divisible by both 3 and 37

111111, 222222, 333333, 444444 …………….. are divisible by 3, 7, 11, 13 37

Test of Divisibility

Divisibility by 2

A number is divisible by 2 if the unit digit is zero or divisible by 2

Example: 12, 26, 128, 1240 etc.

Divisibility by 3

A number is divisible by 3 if the sum of digits in the number is divisible by 3

Example: 2553

Here, 2 + 5 + 5+ 3 = 15, which is divisible by 3 hence 2553 is divisible by 3

Divisibility by 4

A number is divisible by 4 if its last two digits is divisible by 4

Example: 2652

Here, 52 is divisible by 4, so 2652 is divisible by 4

Divisibility by 5

A number is divisible by 5 if the units digit in number is 0 or 5

Example: 20, 35, 140, 165 etc.

Divisibility by 6

A number is divisible by 6 if the number is even and sum of digits is divisible by 3

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Example: 4536

4536 is an even number and also sum of digit 4 + 5 + 3 + 6 = 18 is divisible by 3

Divisibility by 7

To check whether a number is divisible by 7 or not first multiply the units digit of

the number by 2 and subtract it from the remaining digits, continue this process. At

the end if the result becomes ‘0’ or ‘7’ then the number is divisible by 7

For E.g. : 3066

306 | 6

- 12 6 × 2 = 12

-----------

29 | 4

- 8 4 × 2 = 8

-----------

2 | 1

- 2 1 × 2 = 2

-----------

0

-----------

Hence 3066 is divisible by 7

Divisibility by 8

A number is divisible by 8 if last three digit of it is divisible by 8

Example: 47472

Here, 472 is divisible by 8 hence this number 47472 is divisible by 8

Divisibility by 9

A number is divisible by 9 if the sum of its digit is divisible by 9

Example: 108936

Here, 1 + 0 + 8 + 9 + 3 + 6 = 27, which is divisible by 9 and hence 108936 is

divisible by 9

Divisibility by 11

A number is divisible by 11 if the difference of sum of digit at odd places and sum

of digit at even places is either 0 or divisible by 11

Example: 1331

The sum of digit at odd places is 1 + 3 = 0

And, the sum of digit at even places is 3 + 1 = 0

And, their difference is 4 – 4 = 0

So, 1331 is divisible by 11

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TEST OF DIVISIBILITY BY 13

Let us take 2067 … Let us truncate the number

2 0 6 | 7

Add 4 × 7 2 8

2 3 | 4

Add 4 × 4 1 6

3 9, which is divisible by 13

Hence, 2067 is divisible by 13



7 7 5 | 2

Subtract 5 × 2 1 0

7 6 | 5

Subtract 5 × 5 2 5

5 1, which is divisible by 17

Hence, 7752 is divisible by 17



4 8 6 | 4

Add 2 × 4 8

--------

4 9 4

4 9 | 4

Add 2 × 4 8

------

5 7

57, which is divisible by 19

Hence, 4864 divisible by 19

10n – 1 is always divisible by 11 for all even values of n.

i.e. 99, 9999, 999999 are all divisible by 11 [If there are even digits only]

We know 112 = 121 Hence 1012 = 10201, 10012 = 1002001

122 = 144 Hence 1022 = 10404, 10022 = 1004004

132 = 169 Hence 1032 = 10609, 10032 = 1006009

We know 212 = 441, 2012 = 40401

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312 = 961, 3012 = 90601

Condition of Divisibility for Algebraic function

An + Bn is exactly divisible by A + B only when n is odd

Example: A3 + B3 = (A + B)(A2 + B2 – AB) is divisible by A + B, also A5 + B5 is

divisible by A + B

An + Bn is never divisible by A – B (whether n- is odd or even)

Example: A3 + B3 = (A + B)(A2 + B2 – AB) is not divisible by A - B

A7 + B7 is also not divisible by A - B

An – Bn is exactly divisible by A- B (whether n- is odd or even)

Example: A2 – B2 = (A- B) (A+ B) so it is divisible by A- B

A3 - B3 = (A - B)(A2 + B2 + AB) so it is divisible by A- B

Sum of n- natural numbers

S = 1 + 2 + 3 + 4 + 5 + ……………………. n

S = 𝒏 (𝒏+𝟏)

𝟐

Sum of squares of first n- natural numbers

S = 12 + 22 + 32 + ……………………. N2

S = 𝒏 (𝒏+𝟏)(𝟐𝒏+𝟏)

𝟔

Sum of cubes of first n- natural numbers

S = 13 + 23 + 33 + 43 + ……………….. n3

S = [𝒏 (𝒏+𝟏)

𝟐]𝟐

Sum of first n- odd natural numbers

S = 1 + 3 + 5 + 7 + ………………… (2n -1)

S = n2

Sum of first n- even natural numbers

S = 2 + 4 + 6 + 8 + ………………… 2n

S = n2 + n

We should not forget

1 is neither prime nor composite.

The lowest prime number is 2.

The one and only prime number which is even is 2.

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Two consecutive numbers which are prime are only 2 and 3.

The lowest composite number is 4.

There are 4 prime numbers each between 1 and 10 and between 10 and 20.

4. The total number of

One digit numbers are 9 (excluding zero)

Two digit numbers are 90

Three digit numbers are 900

Four digit numbers are 9000

√12 + √12 + √12 + ⋯ α = 4 (answer)

Here, we should find factors for 12 with a difference of 1.

12 = 4 × 3, the answer is 4

√12 − √12 − √12 … … α = 3 (answer = 3)

If the sign is −, the answer is 3 (As 12 = 4×3)

√12√12√12 … … n times = 12𝟐𝒏−𝟏

𝟐𝒏

√12√12√12 … … α = 12 (answer = 12)

DIVIDEND = DIVISOR × QUOTIENT + REMAINDER

DIVISOR ) DIVIDEND ( QUOTIENT

-------------

Remainder

12 ) 170 ( 14

168

------

2

170 = 12 × 14 + 2

ARITHMETIC PROGRESSION (A. P)

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In Mathematics, an Arithmetic Progression or Arithmetic Sequence is a

sequence of numbers such that the difference between the consecutive terms is

constant

For example, the sequence 5, 8, 11, 14, 17, 20 ……… is an Arithmetic Progression

with common difference of 3

The general form of Arithmetic Progression is

a, a + d, a + 2d, a + 3d, a + 4d, …………….

First term = a

Common difference = d

No. of terms = n

Any particular term or n – th term = an

Sum of 1st ‘n’ terms = Sn

an = a + (n – 1) × 𝑑

Sn = 𝑛

2 × [2𝑎 + (𝑛 − 1) × 𝑑]

Or

Sn = 𝑛

2 × ( 𝑎 + 𝑎𝑛)

GEOMETRIC PROGRESSION

In mathematics, a Geometric Progression or Geometric Sequence is a sequence of

numbers where each term after the first is found by multiplying the previous one

by a fixed , non – zero number called the Common ratio

For example, the sequence 2, 6, 18, 54, …………is a Geometric Progression with

Common ratio 3

The general form of a Geometric Progression is,

a, ar, ar2, ar3, ar4, …………………..

First term = a

Common ration = r

Any particular term or n – th term = an

Sum of first n – terms = Sn

an = a × 𝑟𝑛−1

Sn = 𝑎 ×(𝑟𝑛 −1)

𝑟−1 r > 1

Sn = 𝑎 ×(1− 𝑟𝑛)

1−𝑟 r < 1

Ramanujan’s Number (1729)

It is a very interesting number, it is the smallest number expressible as the sum of

two cubes in two different ways

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1729 = 13 + 123 = 93 + 103

1 If the expressions x + 809436 × 809438 be a perfect square, then the value

of ‘x’ is …………..

SOLUTION:

Given that, x + 809436 × 809438

= x + (809437 – 1) × (809437 +1)

= x + (809437)2 – (1)2

So, x should be 1 (answer)

2 The first term of a Geometric Progression (G.P) is 1. The sum of the third

and fifth terms is 90. Find the common ratio of the G.P

SOLUTION:

Given that, first term of G.P (a) = 1

∴ ar2 + ar4 = 90 [from question]

1 × r2 + 1 × r4 = 90 [ a = 1]

By solving, we get

r = ± 3 (answer)

3 A number when divided by by 56, the remainder is 29. If the number is

divided by 8, then the remainder is ………..

SOLUTION:

When first divisor is exactly divisible by second divisor, then the second remainder

is obtained by dividing the first remainder by the second divisor.

Therefore, on dividing 29 by 8, the remainder = 5 (answer)

4 The sum of two numbers is equal to 200 and their difference is 25. The

ratio of the two numbers is ……

SOLUTION:

Let the numbers be a and b

According to question

a + b = 200 …………. (1)

a – b = 25 ……..……. (2)

By equation (1) and (2)

𝑎 + 𝑏

𝑎 − 𝑏=

200

25

𝑎 + 𝑏

𝑎 − 𝑏=

8

1

Using Componemdo and Dividendo method

𝑎 + 𝑏 + 𝑎 − 𝑏

𝑎 + 𝑏 − 𝑎 + 𝑏=

8 + 1

8 − 1

𝑎

𝑏=

9

7 (𝑎𝑛𝑠𝑤𝑒𝑟)

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5 If 𝒂𝟏

𝟑 = 𝟏𝟏 then the value of a2 – 331a is …………..

SOLUTION:

Given that, 𝑎1

3 = 11

a = 113

Therefore, a2 – 331a = a(a – 331)

= 1331(1331 – 331)

= 1331000 (answer)

6 If 1. 5a = 0. 04b, then 𝒃−𝒂

𝒃+𝒂 is equal to …….,

SOLUTION:

Given that, 1. 5a = 0. 04b

𝑏

𝑎=

1.5

0.04=

150

4

By using Componendo and Dividend

𝑏 − 𝑎

𝑏 + 𝑎=

150 − 4

150 + 4

= 73

77 (𝑎𝑛𝑠𝑤𝑒𝑟)

7 The number obtained by interchanging the two digits of a two digit

number is lesser than the original number by 54. If the sum of the two digits

of the number is 12, then what is the original number ………,

SOLUTION:

Suppose required number = 10x + y, where x > y

According to question

(10x + y) – (10y + x) = 54

x – y = 6 …………… (1)

And, x + y = 12 ………… (2)

By solving, equations (1) & (2), we get

x = 9

x + y = 12 => 9 + y = 12

y = 3

∴ Required number = 10 × 9 + 3

= 93 (answer)

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fdaytalk · fdaytalk.com author: j maha laxmaiah mail id: [email protected] 11 × 13 = 143 12...

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