Method 2

Product of three digit numbers

The method of ' Urdhva Tiryagbhyam ' can also be used for findiing the product of three digit numbers . This although a bit difficult from the previous one can work effectively after a small amount of practice.Explaining the method by an example . This methodonly looks long but is very small if compared to the normal method of multipling three digtit numbers . It can be completed in one single line only instead of four lines of normal method .


Example 1 : - Find the product of 234 * 342 .

Answer : -

Step 1 : -

Multiply the last two digit of the multiplier and the multiplicand.
i.e.
4 * 2 = 8

Obtained digit = 8 Carry = 0
This is the right most digit of the answer. If any carry is produced then it is taken to the next step . Like if 18 is produced then 1 is taken to the next step .

Step 2 : -

Coss Multiply the last two - two digits of the multiplier and the multiplicand .

Multiply last digit of multiplicand(23 4) with last second digit of multiplicand(342)
i.e.
4 * 4 = 16
and
Multiply last second digit of multiplier(234) with the ladt digit of multiplicand(342)
i.e.
3 * 2 = 6

Add the two digits and add carry from the fiirst step.
i.e.
16 + 6 + 0 =22
place the right most digit to left of digit obtained in first step and use rest for carry.
Obtained digit = _______28 Carry = 2


Step 3 : -

Now cross multiply the first and the last digit of the multiplicand and multiplier
(Multiplicand) 234 (Multiplier)342
2 * 2 = 4
and
234 342
4 * 3 = 12
also multiply the middle digits
234 342
3 * 4 = 12
Add the above three ovtained digits and carry from the last step
i.e.
4 + 12 + 12 + (Carry)2 = 30
Place right most digit in obtained digit and rest for carry to next step.
Obtained digit = 028 Carry = 3

Step 4 : -

Now cross multiply the first two - two digits of the multiplier and multiplicand.
(Multiplicand) 234 (Multiplier)342
2 * 4 = 8
(Multiplicand) 234 (Multiplier)342
3 * 3 = 9
Add the two digits + the carry from above step
i.e.
8 + 9 + (Carry)3 = 20
Place right most digit in obtained digit and rest for carry to next step.
Obtained Digit = ________0028 Carry = 2

Step 5 : -

Now multiply the first two digits of multipler and multiplicand
234 342
i.e. 2 * 3 = 6
Add the carry from the last step
6 + 2 = 8
Place it to the left and here cames the answer.........

234 * 342 = 80028

One line approach of method 2

One Line Approach
234
342
8

2 3             4
3 42
4*4=16 + 3*2=6 + (carry) 0 =22

2       3       4

3       4        2
12 +12 + 4 + (carry) 2 = 30

2    3    4

3    4    2
9 + 8 + (carry) 3 = 20

      2 34
3 42 Answer = 60028 (Starting from down)
       6

Algebraic Proof : -

Let there be two numbers (ax ² + bx + c) and (dx² + ex + f) . If we have to multiply them then
(a.x² + b.x + c)(d.x² + e.x + f) =
ad.x4 + (bd + ae).x3 + (cd + be + af).x2 + (ce + bf)x + cf

From above we can see that in the expression the cofficients of various terms are the same as described in the various steps .

Note : - We can also start the same procedure from the left most digit also.

Exersises : -

Find the product of the following numbers : -

1.) 230 * 453
2.) 321 * 546
3.) 342 * 786
4.) 467 * 598
5.) 104 * 599