*In this post, we will learn to multiply two numbers whose ending digit is 5 and the sum of the other digits of both numbers is an odd number. For example,*

*Empty space*

**1) 25 * 75 (Both the numbers end in 5 and the sum of the other digits of the numbers is an odd number, that is, (2 + 7) = 9)**

**2) 45 * 95 (Ends in 5 & (4 + 9) = 13)**

**3) 85 * 115 (Ends in 5 & (8 + 11) = 19)**

**4) 135 * 165 (Ends in 5 & (13 + 16) = 29)**Empty space

*Let us consider the first example and see the steps for the actual multiplication.*Empty space

*1) 25 * 75*Empty space

*a) The answer has two parts LHS and RHS.*

*b) The answer looks like LHS/RHS.*

*c) The value of RHS is always 75. That is, RHS = 75.*

*d) To arrive at the value of the LHS, first we need to multiply the other digits of the numbers besides the ending digit 5. That is, in this case, it is 2 * 7 = 14.*

*e) Next, we have to halve the sum of those two numbers and take the integer part of the answer. That is, (2+7)/2 = 9/2 = 4.5 = 4.*

*f) Add the numbers obtained in step (d) & (e).*

*That is, 14 + 4 = 18.*

*g) The final answer is LHS/RHS = 18/75 = 1875.*Empty space

*Solved Examples*Empty space

*2) 45 * 95*

*= ((4*9)+((4+9)/2))/75*

*= (36+(13/2))/75*

*= (36+6)/75*

*=*

*42/75*

*= 4275*

*Empty space*

3) 85 * 1153) 85 * 115

*= ((8*11)+((8+11)/2))/75*

*= (88+(19/2))/75*

*= (88+9)/75*

*= 97/75*

*= 9775*

*Empty space*

4) 135 * 1654) 135 * 165

*= ((13*16)+((13+16)/2))/75*

*= (208+(29/2))/75*

*= (208+14)/75*

*= 222/75*

*= 22275*
Nice method...Really useful..Looking forward to more methods..

ReplyDeleteThanks

Aravindan