Question 782657
The sum of digits of a two digit positive integer is 11. If the digit at one's place is one less than the square of the digit at the ten's place. find the number?
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let u=units digit
let t= tens digit
u+t=11
u=11-t
..
u=t^2-1
11-t=t^2-1
t^2+t-11=0
(t+4)(t-3)=0
t=-4 (reject)
or
t=3
u=t^2-1=9-1=8
number: 38