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 can you write an algebraic equation that represents...

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 can you write an algebraic equation that represents this trick

Asked on by anna101

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

Whenever a number is multiplied by 11 we get as the result a number which has the sum of its odd digits equal to the sum of its even digits. So if you were to multiply any number with 11 and got the result abcd, we have a + c = b + d

Also, when a number is multiplied by 9 the sum of the digits is equal to a multiple of 9.

Here we have 99 which is 9*11. So any number that is a result of multiplying 99 and which lies between 1 and 100 gives a result abcd where a + c = b + d

So we have a + c= b + d and a + b + c + d is a multiple of 9.

=> 2*(a + c) is a multiple of 18. Now, two digits can have a maximum sum of 18, so a + c <=18.

Therefore a + b + c + d is equal to 18.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

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