Question 285172: I am a 2 digit number. The difference between the product of me and myself and the sum of me and myself is 99. What number am i? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=tens digit
And let y=units digit
The number then is 10x+y
Product of me and myself=(10x+y)*(10x+y)
The sum of me and myself=(10x+y)+(10x+y)=2(10x+y)
So our equation to solve is:
(10x+y)^2-2(10x+y)=99 subtract 99 from each side
(10x+y)^2-2(10x+y)-99=0 quadratic in standard form and it can be factored
{(10x+y)-11}{(10x+y)+9}=0
10x+y=11---------------------answer
10x+y=-9-----------------no good, not a 2 digit number
So the 2 digit number is 11
CK
11*11=121
11+11-22
121-22=99
99=99
Hope this helps---ptaylor
