Here is a math trick: If you multiply any digit from 1-100 by 99 and you add the answer you will get 18 all the time. Why does it work.And please...

Any integer number from 1 to 99 is a two digit number ,
could be written like 10x+y, where x is  a digit in 10's place and y is in units
place.

The digits x and y as given in this example should
be such that 0 < x, y < = 9. Or x and y can not be the digit zero. x and y
could be any one of the digits 1,2,3,..9.

Therefore the
product 99(10x+y) = (100-1)(10x+y).

Therefore 99(10x+y) =
(1000x+100y) - (10x+y)

The right side = 1000x +100y -10x-y
=

(1000x +100y +10*0 + 1*0) - 10x -
y.

In decimal format this is written digit by digit like:
(xy00) -(xy) or something like 5600 - 48 = 5*1000 +(6-1)100+(10-1-4)+(10-8) =
5552.

So in actual subtraction algorithm  under such cases
, we borrow 1 from 100th place to 10's place. Now  1 hundred borrowed in 10's place  is
worth 10 tens. Then we we borrow 1 ten from 10's place to units place. Now 1 ten
borrowed is worth 10 in units place.

Therefore 99(10x+y)
=1000x + 100(y-1) + 10*0 +1*0) -10x - y.

99(10x+y)= 1000x
+100(y-1)+10(10-1) +1*10 -10x - y

99(10x+y) =
1000x+100(y-1)+10(10-1-x) +1*(10-y).

From right side we
notice that  digit x is in 1000 's place ,  the digit (y-1) is in 100's place , the
digit (10-1-x ) is in 10's place and (10-y) is the digit in unit's
place.

Therefore the total of the digits in the product
99*(10x+y) = x+(y-1)+(10-1-x)+(10-y ) = x+y-1 +10-1-x+10-y = 20-2 =
18.

Therefore , the product of 99*(any integer from 1 to
99) has the sum of digits 18.

Hope this
helps.