# Multiplying Number

When a 2-digit number is to be multiplied with a two-digit number the following process would be applied. If there were two numbers AB and CD then there product would be calculated as under.

AB

CD

**Step 1: **BD (Write only the unitâ€™s digit and carry the rest to the next step).

**Step 2: **AD + BC + Carry over (Cross multiply and Add, write a single digit and carry the rest to the next step).

**Step 3: **AC + Carry over (Write the complete number because this is the last step).

29

53

**Step 1:** 9Ã—3=27 (Write 7 and 2 is carried over to the next step).

**Step 2:** 2Ã—3+9Ã—5+2(Carried Over)=53 (Write 3 and 5 is carried over to the next step)

**Step 3:** 2 Ã— 5 + 5 (Carried Over) = 15 (Write 15 because this is the last step)

Therefore 1537 is the answer

37

73

**
Step 1: **7Ã—3=21 (Write 1 and 2 is carried over to the next step)

**Step 2: **3Ã—3+7Ã—7+2 (Carried Over)=60 (Write 0 and 6 is carried over to the next step)

**Step 3: **3 Ã— 7 + 6 (Carried Over) = 27 (Write 27 because this is the last step).

Therefore 2701 is the answer.

Try Multiplying 23 and 32 and see if the answer is 736.

Try Multiplying 28 and 82 and see if the answer is 2296.

Now we will try multiplying a three-digit number by a three-digit number. Because there are six digits, the total number of steps would be 5.

ABC

DEF

**Step 1:** CF (Write only the unitâ€™s digit and carry the rest to the next step)

**Step 2:** BF + CE + Carried Over (Write only the unitâ€™s digit and carry the rest to the next step)

**Step 3:**

AF + CD + BE + Carried Over (Write only the unitâ€™s digit and carry the rest to the next step)

**Step 4: **AE + BD + Carried Over (Write only the unitâ€™s digit and carry the rest to the next step)

**Step 5:** AD + Carried Over (Write the complete number because this is the last step)

123

456

**Step 1:** 3Ã—6=18(Write 8 and 1is carried over to the next step)

**Step 2:** 2Ã—6+3Ã—5+1(Carried Over)=28(Write8and2iscarriedover to the next step)

**Step 3:** 1Ã—6+3Ã—4+2Ã—5+2(Carried Over)=30(Write 0 and 3 is carried over to the next step)

**Step 4:** 1Ã—5+2Ã—4+3(Carried Over)=16(Write 6 and 1 is carried over to the next step)

**Step 5:** 1 Ã— 4 + 1 (Carried Over) = 5 (Write 5 because this is the last step). Therefore 56088 is the answer.

243

172

**Step 1:** 3 Ã— 2 = 6 (Write 6 which is the single digit number)

**Step 2:** 4Ã—2+7Ã—3=29(Write 9 and 2 is carried over to the next step)

**Step 3:** 2 Ã— 2 + 1 Ã— 3 + 4 Ã— 7 + 2 (Carried Over) = 37 (Write 7 and 3 is carried over to the next step)

**Step 4:** 2 Ã— 7 + 4 Ã— 1 + 3 (Carried Over) = 21 (Write 1 and 2 is carried over to the next step)â€¨

**Step 5:** 2 Ã— 1 + 2 (Carried Over) = 4 (Write 4 as this is the last step)

Therefore 41796 is the answer.

# Eliminating choice

If we are required to check multiplication of 2734 Ã— 6034, then the result will have the last digit the same as the digit of the product of 4 and 4. (in this case 6)

**Multiplication of 4 digit nos**

Always focus on the 2 left-most digits to get approx answer. eg. 2734 Ã— 6034 is approx = 27Ã—60 =1620,00, 00

**Squaring Numbers**

**Part â€“ I: **While squaring a number, you need a base. All those numbers can be taken as bases, which have a 1 and the rest number of zeroes with them (i.e. the complete round numbers like 100, 1000, 10000 etc.). The square of a number will have two parts, the left part and the right part. There is no limit for the left side, but the right side must have as many digits as the number of zeroes in the base i.e. if 100 is taken as base there should be 2 zeroes on the right side and if 1000 is taken as base then the number of digits on RHS should be 3.

Now take a number 92. The nearest complete base is 100. The difference between the base and the number given is 8. The square of this difference is 64, which will become the right side. Q It is already having two digits, so it would be simply placed on the right side. Now the difference of 8 is subtracted from the number given i.e. 92 â€“ 8 = 84 and it will become the left side. Therefore the square of 92 is 8464.

Take another number say 94, which is 6 less than the base. While squaring 94, the right side will be (6)2 i.e. 36. And the left side would be the number given â€“ difference i.e. 94 â€“ 6 = 88. So the square of 94 is 8836. If the square of the difference is having lesser digits then required, then in order to have the needed number of digits on the right side, 0â€™s can be put with the square. e.g. if you square a number like 97, difference is 3. The right side in this case would become 09, because 9 is a single digit number and youâ€™ll have to put a â€˜0â€™ before it to make it a two-digit right hand side. The left side would be 97 â€“ 3 = 94. The square is 9409.

In case, the number of digits is more than needed, then the extra digits are carried to the left side. e.g. take 86. The difference is 14 and the square of the difference is 196, which is a 3-digit number, so the 3rd extra digit 1 would be carried to the left side. The left side is 86 â€“ 14 = 72 + 1â€¨= 73. (carried over).

So the square of the number is

You can practice the following squares to get expertise on squaring.

87 _____â€¨96 _____â€¨82 _____

85 _____ 97 _______ 99 _____ 81 _____

79 ______ 77 _____ 91 _______ 88 _____

91 _____ 92 ______ 76 _____ 89 _______

**Part â€“ II :**

If the number to be squared is greater than the base, then there is only one difference in approach, i.e. the difference between the number and the base is to be added in the number instead of subtracting. Take a number 107. The difference is 7. The right side will have square of difference i.e. (7)2 = 49. And the left side will be 107 + 7 = 114 Q the number is greater than the base. So the square is 11449.

Similarly even in this case, if the number of digits on the right side less than the required number, then you can write â€˜0â€™s with it to get the right side. The square of 103 would be : the difference is 3, its square is 9, which is a single digit number, so a 0 would be written with it i.e. 09. Then the left side is 103 + 3 = 106. The square becomes 10609.

In case the square of the difference is a 3-digit number, then the third digit would be carried to the left side. Consider one number say 118. The difference is 18 ⇒ (18)2 ⇒ 324. Out of this 3-digit number the third digit 3 would be taken to the left side. The left side would become 118 + 18 + 3 (Carried) = 139 and the square would be

136- - + - - 324** ****13924**

# Cyclicit

This concept is useful to find the last digit of a number raised to a high power:

Cyclicity has been explained in the previous chapter.

What is the last digit of 723 Ã— 813?

Divide 23 by 4, we get remainder 3. Hence last digit of 723 will be same as 73 or 3.

Divide 13 by 4, we get remainder as 1, hence last digit of 813 is same as 81 or 8.

So the last digit of the product is 3 Ã— 8, or 4.