Introduction:

*** Multiplicand:** The number to which multiplication is done is called multiplicand.

*** Multiplier:** The number that is multiplied is called multiplier.

*** Multiplication: **The product of multiplier and multiplicand is called multiplication. Hence multiplicand × multiplier = multiplication

Example: 215 × 20 = 4300

The short-cut method of multiplication can be divided into two parts.

1. General Rules: By this method, the product of all types of numbers can be extracted.

2. Specific Rules: This method allows the product of certain types of numbers quickly.

Noble Suggestion: In Bank Clerical exams more questions related to the simplification are asked. Multiplication is very important in solving such questions. Therefore, in order to solve more questions in less time, it is very important to adopt the Short-Cut Methods of multiplication to get the product of numbers. Also remember the Table of 1-30, so that they can multiply quickly.

Type – 1: Multiplication of two, two-digit numbers:

Trick :

General Form: AB × CD =?

Working Steps:

1st Step. B × D

2nd Step. A × D + B × C

Last Step. A × C

Example: 42 × 56 =?

Explanation:

1st Step. (2 × 6) = 12 ⇒ _{ 1}2

2nd Step. (4 × 6 + 2 × 5) = 34 ⇒ _{ 3}4

Last Step.(4 × 5) = 20

Expected product = 20 _{ 3}4 _{1}2 = 2352 Answer.

Note: Here are numbers 1 and 3 are obtained from carry of 12 and 34 respectively.

** Type – 2: Multiplication of two, three-digit numbers: **

Trick:

General Form: ABC × DEF =?

Working Steps:

1st Step. C × F

2nd Step. B × F + C × E

3rd Step. A × F + B × E + C × D

4th Step. A × E + B × D

Last Step. A × D

Example: 346 × 527 =?

Explanation:

1st Step. 6 × 7 = 42 ⇒ _{ 4}2

2nd Step. 4 × 7 + 6 × 2 = 40 ⇒ _{ 4}0

3rd Step. 3 × 7+ 4 × 2 + 6 × 5 = 59 ⇒ _{ 5}9

4th Step. 3 × 2 + 4 × 5 = 26 ⇒ _{ 2}6

Last Step. 3 × 5 = 15 ⇒15

Expected product = 15_{2}6_{5}9_{4}0_{4}2 = 182342

Note: Here are 4, 4, 5 and 2 numbers obtained from 42, 40, 59 and 26, respectively.

Exercise – Solve them by TRICK –

(1) 234 × 567 =?

(2) 436 × 714 =?

(3) 814 × 532 =?

(4) 427 × 931 =?

(5) 732 × 453 =?

** Type – 3: Multiplication of two, four-digit numbers: **

Trick:

General Form: ABCD × EFGH =?

Working Steps:

1st Step. D × H

2nd Step. C × H + D × G

3rd Step. B × H + C × G + D × F

4th Step. A × H + B × G + C × F + D × E

5th Step. A × G + B × F + C × E

6th Step. A × F + B × E

Last Step. A × E

Example: 5314 × 6272 =?

Explanation: 1st Step. 4 × 2 = 8

2nd Step. 1 × 2 + 4 × 7 = 30 ⇒ _{ 3}0

3rd Step. 3 × 2 + 1 × 7 + 4 × 2 = 21 ⇒ _{ 2}1

4th Step. 5 × 2 + 3 × 7 + 1 × 2 + 4 × 6 = 57 ⇒ _{ 5}7

5th Step. 5 × 7+ 3 × 2 + 1 × 6 = 47 ⇒ _{ 4}7

6th Step. 5 × 2 + 3 × 6 = 28 ⇒ _{ 2}8

Last Step. 5 × 6 = 30

Expected product = 30_{2}8_{4}7_{3}1_{3}0 = 3332940 Answer.

Note: Similarly, the product of five and six digit numbers can also be known.

** Type – 4: Multiplication of two numbers of different lengths: **

1. If any number of three digits is to be multiplied by any other two digit number, then the number of three digits is created by placing a zero (0) on the left of the two digit number and the Multiplication is obtained by multiplying two digit number with the number of three digits.

2. If a number of four digits is to be multiplied by any other number of three digits, then the number of four digits is created by placing a zero (0) on the left of the three digit number and Multiplication is done by multiplying the number of three digits with the number of four digits.

Example 1. 753 × 46 =?

Explanation: 1st Step. 3 × 6 = 18 ⇒ _{ 1}8

2nd Step. 5 × 6 + 3 × 4 = 42 ⇒ _{ 4}2

3rd Step. 7 × 6 + 5 × 4 + 3 × 0 = 62 ⇒ _{ 6}2

Last Step. 7 × 4 +5 × 0 = 28 ⇒ _{ 2}8

Expected product = 28_{6}2_{4}2_{1}8 = 34638 Answer.

Example 2. 4326 × 532 =?

Explanation: 1st Step. 6 × 2 = 12 ⇒ _{ 1}2

2nd Step. 2 × 2 + 6 × 3 = 22 ⇒ _{ 2}2

3rd Step. 3 × 2 + 2 × 3 + 6 × 5 = 42 ⇒ _{ 4}2

4th Step. 4 × 2 + 3 × 3 + 2 × 5 + 6 × 0 = 27 ⇒ _{ 2}7

5th Step. 4 × 3 + 3 × 5 + 2 × 0 = 27 ⇒ _{ 2}7

Last Step. 4 × 5 + 3 × 0 = 20

Therefore, the desired product = 20_{2}7_{2}7_{4}2_{2}2_{1}2 = 2301432 Ans.

Exercise – Solve them by TRICK

(1) 367 × 56 =?

(2) 432 × 92 =?

(3) 64 × 327 =?

(4) 4326 × 538 =?

(5) 7432 × 627 =?

(6) 234 × 8321 =?

(7) 53261 × 234 =?

(8) 4326 × 54213 =?

(9) 4325 × 36 =?

** Type – 5 **

** Multiplication in a Certain Number by Any Number from 11 to 19: **

Trick: To multiply any number from 11 to 19 in any multiplication multiplied by the multiplier of each of that multiplication.

Example: 34526 × 13 =?

Explanation: 1st Step. 6 × 13 = 78 ⇒ _{ 7}8

2nd Step. 2 × 13 = 26 ⇒ _{ 2}6

3rd Step. 5 × 13 = 65 ⇒ _{ 6}5

4th Step. 4 × 13 = 52 ⇒ _{ 5}2

Last Step. 3 × 13 = 39

Therefore, the desired product = 39_{5}2_{6}5_{2}6_{7}8 = 448838 Answer.

Exercise – Solve them by TRICK

(1) 4153 × 11 =?

(2) 57324 × 12 =?

(3) 673216 × 13 =?

(4) 93245 × 14 =?

(5) 23456 × 15 =?

(6) 34326 × 16 =?

(7) 123458 × 17 =?

(8) 41237 × 18 =?

(9) 72314 × 19 =?

** Specialty of Multiplication on Whole Numbers: **

** Type – 1 same and Ten rules: **

If the product of such two numbers is to be known, whose sum of digits is ’10’ and the remaining numbers are equal, then by multiplying the digits of the unit by multiplying both points (if the product comes in one digit, just before it) A zero ‘(0) is completed and two digits are completed) on the right side is placed and by adding “1” in the same number, multiplying the number with the same number, Keeping the right (left) on the right, the product of these numbers gets achieved.

(Here x is the same number and y & z respectively are the units of the unit digit of both numbers)

Example: 86 × 84 =?

Explanation: 86 × 84 = 8 × (8 + 1) / 6 × 4 = 7224

Exercise – Solve them by TRICK

(1) 35 × 35 =?

(2) 97 × 93 =?

(3) 206 × 204 =?

(4) 129 × 121 =?

(5) 1308 × 1302 =?

**Type – 2 Same and Five Rule **

If the product of two numbers is to be calculated, whose unit digits have the sum ‘5’ and the remaining numbers are equal, then a zero (0) is placed on the left side after multiplying unit digits. After that, by adding a ‘1/2’ part of the same number in square of the same number, the multiplication of those numbers obtained by keeping the number before zero (‘0’). Thus multiplication =

(Here x is same number and y,z respectively are unit numbers of both numbers.)

Note: If ‘1/2’ is in the sum of the ‘square’ + ‘1/2’, then add ‘5’ in the zero at the position of ‘ten’ for the ‘1/2’.

**Type – 3 Ten and Same Rule **

Trick: If the unit digits are identical and the sum of the tenth digit is 10, then the number obtained by the square of the digits of the unit (if the single digit is received after the square, then a zero (0) is placed on the left side of it)placed on the right side and the unit digit is added in the sum of tenth digit and resultant number is placed on the left side.

Thus, the product = y × z + x / x^{2} (Here y and z are respectively ten digit of first and second numbers and z is the same number of unit)

**Type – 4: Five and Same Rule**

Trick: If the unit digits of two numbers is identical and the sum of the tenth digit is ‘5’ then the two digits of the number obtained by the square of the unit digit of same number (If the number obtained after squaring the same number is in one digit, then a zero (0) is placed on its left side) is placed on the right side and in the product of tens digits, half ‘1/2’ of the unit number is added and place on the left side of both numbers.

Thus multiplication =

(Here y and z is the number of tenths of first and second numbers respectively)

Note: 1/2 = 10/2 = 5, that is, for ½, 5 is added to the number 4 on the right.

**Type – 5 Same and Twenty Rule **

Trick : If the sum of last two digits of multiplicand and the last two digits of multiplier are 20 and the remaining numbers are equal, then four digits of the product of the last two digits of multiplicand and the the last two digits of multiplier ( If not in the four digits, the left digit is filled with zero (0)) placed on the right side and add 1/5 of the same number in the square of same number and place them on the left side. the product of these numbers is computed. Thus multiplication =

(Here x is the same number and y,z are respectively two-two digits of multiplier and multiplicand.)

Note: 1/5 = 10/5 = 2, that is, for 1/5, 2 is added to zero (0) on the right side.

Exercise – Solve them by TRICK

(1) 4506 × 4514 =?

(2) 2411 × 2409 =?

(3) 2504 × 2516 =?

(4) 1815 × 1805 =?

(5) 1503 × 1517 =?

** Type – 6 Same and Forty Rule: **

Trick : If the sum of the multiplicand and the multiplier is equal to 40, and the remaining numbers are identical, then the four points of multiplication of the last two digits of the multiplicand and multiplier (if the product is not in four digits, the four digits is completed by placing zero (0) on the left side) is placed on the right side and by adding 2/5 of the same number in square of the same number, keeping them on the right side, the product of these numbers Will be the answer. Thus multiplication =

(Here x is same number and y,z are respectively two digits of the multiplicand and multiplier.)

Note: 4/5 = 40/5 = 8 that is, for 4/5, 8 is added to zero (0) on the right side.

Exercise – Solve them by TRICK –

(1) 5523 × 5517 =?

(2) 1512 × 1528 =?

(3) 2231 × 2209 =?

(4) 3618 × 3622 =?

(5) 1308 × 1332 =?

** Type – 7 Same and Fifty Rule **

Trick : If the sum of the multiplicand and the multiplier is equal to 50, and the remaining numbers are identical, then the four points of multiplication of the last two digits of the multiplicand and multiplier (if the product is not in four digits, the four digits is completed by placing zero (0) on the left side) is placed on the right side and by adding 1/2 of the same number in square of the same number, keeping them on the right side, the product of these numbers Will be the answer. Thus multiplication =

(Note: Here x is the same number and y,z respectively are two-two digits of multiplicand and multiplier.)

Note: 1/2 = 10/2 = 5, that is, for ½, 5 is added to the zero (0) on the right side.

Exercise – Solve them by TRICK –

(1) 4032 × 4018 =?

(2) 2428 × 2422 =?

(3) 2919 × 2931 =?

(4) 2535 × 2515 =?

(5) 934 × 916 =?

** Type – 8 Same and Sixty Rule:**

Trick : If the sum of the multiplicand and the multiplier is equal to 60, and the remaining numbers are identical, then the four points of multiplication of the last two digits of the multiplicand and multiplier (if the product is not in four digits, the four digits is completed by placing zero (0) on the left side) is placed on the right side and by adding 3/5 of the same number in square of the same number, keeping them on the right side, the product of these numbers Will be the answer. Thus multiplication =

(Here x is the same number and y and z respectively 2-2 digits of multiplicand and multiplier.)

Note: 1/5 = 10/5 = 2 that is for 1/5, 2 is added to zero (0) on the right side .

** Type – 9 Same and Eighty Rule **

Trick : If the sum of the multiplicand and the multiplier is equal to 80, and the remaining numbers are identical, then the four points of multiplication of the last two digits of the multiplicand and multiplier (if the product is not in four digits, the four digits is completed by placing zero (0) on the left side) is placed on the right side and by adding 4/5 of the same number in square of the same number, keeping them on the right side, the product of these numbers Will be the answer. Thus multiplication =

(Note: x is the same number and y and z respectively two-two digits of multiplicand and the multiplier)

Note : 3/5 = 30/5 = 6 that is, for 3/5, 6 is added to the number 1 on the right side.

Exercise – Solve them by TRICK

(1) 565 × 515 =?

(2) 832 × 848 =?

(3) 4546 × 4534 =?

(4) 5225 × 5255 =?

(5) 1645 × 1615 =?

** Type – 10 Same and Hundred Rule**

Trick :

If the sum of the multiplicand and the multiplier is equal to 100, and the remaining numbers are identical, then the four points of multiplication of the last two digits of the multiplicand and multiplier (if the product is not in four digits, the four digits is completed by placing zero (0) on the left side) is placed on the right side and by adding the same number in square of the same number, keeping them on the right side, the product of these numbers Will be the answer. Thus multiplication = x^{2 }+ x / y × z

(Note: Here x is the same number and y and z are respectively two-two digits of multiplicand and multiplier.)

Exercise Solve them by TRICK

(1) 1773 × 1727 =?

(2) 2956 × 2944 =?

(3) 3678 × 3622 =?

(4) 6585 × 6515 =?

(5) 8268 × 8232 =?

** Multiplication of any number by Repeated Digit Numbers **

Type – 1

Trick : If the number of four digits is to be multiplied by ’11’, then the sum of the digits of the unit number, unit and tenth digit of the number, the sum of the tens and the hundredths, the hundredth and the thousand and finally the thousand points are placed at the place of unit, ten, hundredth, thousand and ten thousand, respectively for product, as well as added to their left.

Note: Here the multiplier 11 has two digits, so maximum two digits are added. Similarly, 111 has three digits, so maximum three digits will be added and similar action is done with ‘1111’, ‘11111’ etc.

Example 1. 2346 × 11 =?

Explanation: 1st Step. = 6

2nd Step. (4 + 6) = 10 ⇒ _{ 1}0

3rd Step. (3 + 4) = 7

4th Step. (2 + 3) = 5

Last Step. 2

Therefore, the desired product = 257_{1}06 = 25806 Answer.

Example 2. 4327 x 111 =?

Explanation : 1st Step. = 7

2nd Step. (2 + 7) = 9

3rd Step. (3 + 2 +7) = 12 ⇒ _{ 1}2

4th Step. (4 + 3 + 2) = 9

5th Step. (4 + 3) = 7

Last Step. 4

Therefore, the desired product = 479_{1}297 = 480297 Answer.

Example 3. 53148 × 1111 =?

Explanation: 1st Step. 8

2nd Step. (4 + 8) = 12 ⇒ _{ 1}2

3rd Step. (1 + 4 + 8) = 13 ⇒ _{ 1}3

4th Step. (3 + 1 + 4 + 8) = 16 ⇒ _{ 1}6

5th Step. (5 + 3 + 1 + 4) = 13 ⇒ _{ 1}3

6th Step. (5 + 3 + 1) = 9

7th step. (5 + 3) = 8

Last Step. 5

Therefore, the desired product is 589_{1}3_{1}6_{1}3_{1}28 = 59047428 Ans.

Note: If the number of digits of multiplicand is less than the number of digits of the multiplier, then by placing zero (0) on the left side of the multiplicand, the number of digits of the multiplier is equalized and obtained the product as per above rule goes.

Example 4. 734 × 1111 =?

Explanation: ? = 0734 × 1111

1st Step. 4

2nd Step. (3 + 4) = 7

3rd Step. (7 + 3 + 4) = 14 ⇒ _{ 1}4

4th Step. (0 + 7 + 3 + 4) = 14 ⇒ _{ 1}4

5th Step. (0 + 7 + 3) = 10 ⇒ _{ 1}0

Last Step. (0 + 7) = 7

Expected Product = 7_{1}0_{1}4_{1}474 = 815474 Answer

Exercise – Solve them by TRICK –

(1) 234 × 11 =?

(2) 4563 × 111 =?

(3) 632 × 111 =?

(4) 52362 × 1111 =?

(5) 7324 × 1111 =?

(6) 32145 × 11111 =?

(7) 823 × 1111 =?

(8) 3267 × 11111 =?

(9) 431 × 11111 =?

(10) 1234 × 111111 =?

** Type – 2 Multiplication By Repeated digit 2 to 8 Numbers **

Trick : If a four-digit number is to be multiplied by 222, then the sum of the digits of the unit of that number, the sum of the units and the tens, the sum of the unit, the tens, and the hundreds, The sum of the tens, hundred and thousand digits and multiplying by 2, and the result would placed on units, ten, hundredth, thousandth, ten thousand and lakhs respectively. At the same time carry is added to left side.

Note:

(i) If the multiplier is two, four or five digits then the sum of the maximum digits will be two, four and five digits, respectively.

(ii) If the revision number 3, 4, 5 … or 8, the appropriate position will be multiplied by 3, 4, 5. or 8, respectively.

Example 1. 6324 × 222 =?

Explanation:

1st Step. 4 × 2 = 8

2nd Step. (2 + 4) × 2 = 12 ⇒ _{ 1}2

3rd Step. (3 + 2 + 4) × 2 = 18 ⇒ _{ 1}8

4th Step. (6 + 3 + 2) × 2 = 22 ⇒ _{ 2}2

5th Step. (6 + 3) × 2 = 18 ⇒ _{ 1}8

Last Step. 6 × 2 = 12

Therefore, the desired product = 12_{1}8_{2}82_{1}8_{1}28 = 1403928 Ans.

Example 2. 234 × 33 =?

Explanation: 1st Step. 4 × 3 = 12 ⇒ _{ 1}2

2nd Step. (3 + 4) × 3 = 21 ⇒ 21

3rd Step. (2 + 3) × 3 = 15 ⇒ _{1}5

Last Step. 2 × 3 = 6

Expected product = 6_{1}5_{2}1_{1}2 = 7722 Answer.

Example 3. 1231 × 6666 =?

Explanation: 1st Step. 1 × 6 = 6

2nd Step. (3 + 1) × 6 – 24 ⇒ _{ 2}4

3rd Step. (2 + 3 + 1) × 6 = 36 ⇒ _{ 3}6

4th Step. (1 + 2 + 3 + 1) × 6 = 42 ⇒ _{ 4}2

5th Step. (1 + 2) × 6 = 18 ⇒ _{ 1}8

Last Step. 1 × 6 = 6

Therefore, the desired product = 6_{1}8_{3}6_{4}2_{ 3}6_{2}46 = 8205846 Ans.

Note: If the number of points of multiplicand is less than the number of digits of the multiplier, then by placing zero (0) on the left side of the multiplicand, the number of points of the multiplier is equalized and obtained the product as per above rule goes.

Example 4. 432 × 88888 =?

Explanation: ? = 00432 × 88888 = …

1st Step. 2 × 8 = 16 ⇒ _{ 1}6

2nd Step. (3 + 2) × 8 = 40 ⇒ _{ 4}0

3rd Step. (4 + 3 + 2) × 8 = 72 ⇒ _{ 7}2

4th Step. (0 + 4 + 3 + 2) × 8 = 72 ⇒ _{ 7}2

5th Step. (O + 0 + 4 + 3 + 2) × 8 = 72 ⇒ _{ 7}2

6th Step. (4 + 3) × 8 = 56 ⇒ _{ 5}6

Last Step. 4 × 8 = 32

Therefore, the desired product = 32_{5}6_{7}2_{7}2_{ 7}2_{4}0_{1}6 = 38399618 Answer.

Exercise – Solve them by TRICK –

(1) 326 × 22 =?

(2) 7321 × 333 =?

(3) 2341 × 4444 =?

(4) 516 × 555 = 7

(5) 623 × 6666 =?

(6) 243 × 77777 =?

(7) 824 × 888 =?

(8) 1213 × 88888 =?

** Multiplication of two Numbers in the Neighborhood of 100 **

Type – 1

Trick : If both multiplicand and multipliers are less than 100 numbers, then the numbers of two numbers less than 100, by multiplying those deficiencies, two points of the received number are placed on the right side of the product and the deficiency of first number subtracted from the second number, the received number is placed on the left side for the product.

Note: If the product of the deficiencies comes in one digit, then two digits are completed by placing zero (0) on the left and if the product comes in three digits then one digit on its left is taken as carry.

Example 1. 96 × 92 =?

Explanation: ? = 96 – 8/4 × 8 = 88/32 = 8832 [“100-96” = 4 and “100-92” = 4]

Example 2. 83 × 87 =?

Explanation: ? = 83 – 13/17 × 13 = 70/21 = 7221 Ans. [“100-83” = 17 and “100-87” = 13)

Example 3. 96 × 98 =?

Explanation: ? = 96 – 2/4 x 2 = 94/08 = 9408 Ans. [“100-96” = 4 and “100 – 98” = 2]

** Type – 2 **

Trick : If multiplicand and multipliers are both numbers greater than 100, then the two numbers more than 100, by multiplying the difference, two points of the given product are placed on the right side of the product of the given numbers. And the number obtained by adding the second number difference in the first number is placed on the left for the product.

Note: If the product of the differences comes in one point, then the remaining two digits are completed by placing zero (0) on the left side. If the product comes in three digits then one digit of the left is taken as carry.

Example 1. 107 × 109 =?

Explanation: ? = 107 + 9/7 × 9 = 116/63 = 11663 Answer.

Example 2 103 × 102 =?

Explanation: ? = 103 + 2/3 × 2 = 105/06 = 10506 Answer.

** Type – 3 **

Trick : If one of the multiplicand and multipliers is less than 100 and the second number is greater than 100, then if the numbers increase or decrease by 100 in the two numbers, then from the multiplication of both The two digits are taken from one digit to the left and placed on the right side of the product of the given numbers, and by adding the second number increment in the first number or decreasing it from that number is placed on the right of the product of those numbers.

Example 1. 93 x 108 =?

Explanation: ? = 93 + 8/7 × 8 = 101/56 = 100 / (100 – 56) = 10044 Answer.

Example 2. 86 × 109 =?

Explanation: ? = 86+ 9/14 × 9 = 95/126 = 93 / (200-126) = 93/74 – 9374 Αns.

Note: is 100 to 126 more. Therefore, from the left side, 2 out of 200 is reduced to 126.

Exercise – Resolve it with Trick.

(1) 92 × 97 =?

(2) 88 × 96 =?

(3) 107 × 103 =?

(4) 94 × 107 =?

(5) 91 × 119 =?

** Multiplication of Two Numbers in the Neighborhood of 1000: **

Trick : If both the multiplicand and multiplier numbers are less than 1000 or larger or one of them is smaller and the second number is larger, then the numbers increase by 1000 or so, by multiplying both of them, Three digits of the number are placed on the right of the product of the given numbers, and by reducing the number of second number or the increase in the first number, the number obtained is placed on the left side.

Note: In multiplicand and multiplier, if one number is less than 1000 and the second number is larger, then by subtracting the decrease and the increase, the three digits of the product obtained are subtracted from 1000 and placed on the right side for the product.

Example 1. 994 × 991 =?

Explanation : ? = 994 – 9/6 × 9 = 985/054 = 985054 Answer.

Example 2. 1012 × 1019 =?

Explanation: ? = 1012 + 19/12 × 19 = 1031/228 = 1031228 Answer.

Example 3. 982 × 1014 =?

Explanation: ? = 982 + 14/18 × 14 = 996/252

= 995 / (1000-252) = 995/748 = 995,748 Answer.

Exercise – Solve them by TRICK –

(1) 986 × 983 =?

(2) 1015 × 1021 =?

(3) 1009 × 1013 =?

(4) 978 × 1017 =?

(5) 979 × 1023 =?

** Multiplication of Two Numbers in the Neighborhood of 50 **

Trick : If multiplicand and multiplier both numbers are less than or equal to 50, the product of the given numbers of two digits given by multiplying that growth or deficiency of both numbers will be more or less than 50 to the right. After this the growth of the second number in the first number is added or the difference is reduced. Thus by dividing the received number from 2, the quotient is placed on the left for the product of numbers.

Note : For 1/2 = 10/2 = 5 that is 1/2, 5 is added to the number 2 on the right.

Exercise :

(1) 57 × 56 =?

(2) 52 × 58 =?

(3) 41 × 48 =?

(4) 43 × 47 =?

(5) 42 × 49 =?

** Multiplication. By 5, 25, 125 and 625 **

Trick : If a number is to be multiplied by 5, 25, 125 or 625, then by placing one, two, three or four zeros respectively on the right side of that number, divide it by 2,4, 8 or 16 to get multiplication.

Example 1. 4326 × 5 =?

Explanation: ? = 4326 × 5 = 43260/2 = 21630 Answer.

Example 2. 53624 × 25 =?

Explanation : ? = 53624 × 25 = 5362400/4 = 1340600 Answer.

Example 3. 6237 × 125 =?

Explanation : ? = 6237000/8 = 779625 Answer.

Example 4. 732 × 625 =?

Explanation : ? = 7320000/16 = 457500 Answer.

Exercise – Resolve it with Trick.

(1) 7354 × 5 =?

(2) 3692 × 25 =?

(3) 6725 × 125 =?

(4) 234 × 625 =?

(5) 437 × 625 =?

** Multiplication By 75: **

Trick : If any number is to be multiplied by 75, then place 3 zeroes on the right side of that number, multiplying by three and divided by 4 gives the product.

Example: 3467 × 75 =?

Explanation: ? = 3467 × 75 = 3 × 349799 = 3 × 86675 = 260025 Answer.

** Unit digit 5 and the difference 10 rule **

Trick : If the product of two numbers is to be calculated, whose unit’s digit is 5 and its difference is 10, then on the right side 75 is written for the product of those numbers and the except number 5 of units of both numbers add 1 in the digit which is larger, by multiplying it by the other remaining number, the product obtained is placed to the left for the product of those numbers.

Example : 125 × 135 =?

Explanation : ? = 125 × 135 = 12 × (13 + 1) / 75

= 156/75 = 15675. Answer.

Exercise – Solve them by TRICK –

(1) 4265 × 75 =?

(2) 73216 × 75 =?

(3) 205 × 215 =?

(4) 145 × 155 =?

(5) 315 × 325 =?

** General Rules of Multiplication on Mixed Numbers **

** Type – 1 integral and fractional rules of both Rules **

**Type – 2: Integral parts same and Fractional Parts Different Rule **

**Type – 3: Integral Parts Different and Fractional Parts same Rule **

** Special Rules of Multiplication on Mixed Numbers **

** Type – 1 Same and One rule: **

Trick : If the product of two Mixed Numbers is to be calculated whose sum of Fractional Parts is 1 and the Integral Part is equal, then by multiplying Fractional Parts, the product gets written on the right side And by adding 1 to one of the equal integral part, multiplying the number by equal number, the product is written on the left side.

** Type – 2 Same and Half Rule: **

Trick : If the product of two Mixed Numbers is to be calculated, whose sum of Fractional Parts is ½ and the integral part is equal, then by multiplying Fractional Parts for the product of those numbers, the product gets written on the right side And the sum obtained by adding half (1/2) of it in the square of one of the same integral part is written on the left side for the product of those numbers.

Note: If the sum of the fractional part is 1/3, 1/4, 1/8, 1/16 … and so on in the square of the same Integral part, than integral part is 1/3 1 / 4 1/8 1/16 … is added by multiplying it, and this received sum is written on the left side.

**Multiplication of Mixed Numbers By whole Numbers: **

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