One less than the previous or One less than the before :As the name indicates this sutra involves subtracting one from the given number to get our final result.This sutra is highly helpful in case of multiplication by 9,99,999....to any other number and in solving fractions of certain numbers like (1/7),(1/13),(1/17)..etc.

**PART 1: For solving Multiplications**

**Example 1:**

6 * 9 = ?

Step 1 : Minus one from the number on L.H.S digit above

Step 2 : Minus the answer(result) got from step 1 from R.H.S digit (i.e 5 from number 9)

6-1 = 5

9-5 = 4

Combine them LHS,RHS ,to get 54

Therefore,

**The Result for 6 * 9 = 54****Example 2:**

Now lets try the same with 99

75 * 99 =?

Step 1 : 75 -1 = 74

Step 2 : 99 -74 = 25

Combining the above result, 7425

Therefore,

**The Result : 75 * 99 = 7425****Example 3:**

Now lets try the same again with 999

899 * 999 =?

Step 1: 899-1 = 898

Step 2: 999-898 = 101

Combining the above, 898101

Therefore,

**The Result :899 * 999 = 898101****Example 4:**

Try 9765 * 9999 =?

Step 1 : 9765 - 1 = 9764

Step 2 : 9999 - 9764 = 0235

Therefore,

**The Result : 9765 * 9999 = 97640235***Note:*

This can be done to any number of digits of 9...is'nt it wonderful ! .Wondered why?Thats the Beauty of number '9'.Check here to get the answer.

**:**

__Algebraic Proof__Take any two digit number in form of (10 x + y)

Now,multiplying it with (10x + y ) x 99

= ( 10x + y ) x ( 100 – 1 )

= ( 10x + y ) x ( 100 – 1 )

= 10x . 100 - 10x +100y - y

= 10x . 10

= x . 10

= x . 10

= 10x . 10

^{2}– 10x + 10^{2}.y – y= x . 10

^{3}+ y . 10^{2}– ( 10x + y )= x . 10

^{3}+ ( y – 1 ) . 10^{2}+ [ 10^{2}– ( 10x + y )]Look at our answer.We can that it is a four digit number whose 1000

^{th}place is x, 100^{th}place is ( y - 1 ) and the two digit number which makes up the 10^{th}and unit place is the number obtained by subtracting the multiplicand from 100.Thus in Example 2 ,

75 * 99.

The 1000

The 100

Number in the last two places 100-75=25.

We got the answer as 7425

^{th}place is x i.e. 7The 100

^{th}place is ( y - 1 ) i.e. (5 - 1 ) = 4Number in the last two places 100-75=25.

We got the answer as 7425

Giving,

**75 * 99 =****7425**
Carry on, wizard!

ReplyDeletehow cn it work for 9765*99

ReplyDelete@anonymous - hadn't it been worked out above?.It will work for any digit's of 9 and you can try it yourself

ReplyDelete9765*99=(9765-98)/(9-6)/(10-5)

ReplyDelete=966735

I do not have time to explain

@anonymous

ReplyDeleteThanks for the input.As far as your math is concerned, it should follow some pattern or some algebraic proof for some similar type of numbers.

And as far as I tried it, I could'nt come to any conclusions with this or may be I am not sure.It would be great if you could share your ideas with our readers with more details of the same.

That was...beautiful.

ReplyDelete*sniff*