Question

If a two-digit number multiplied by itself gives 6889 and the sum of its digits is 11, find the number

Answer 1

ANSWER
83

EXPLANATION
This method makes use of a system of equations

Let x be the tens-place digit
Let y be the ones-place digit
(therefore, x and y are positive integers between 0 and 9 inclusive as they represent digits)

Then this two digit number is 10x + y
"Multiplying it by itself gives 6889" implies

   (10x + y)(10x + y) = 6889

or just

   (10x + y)² = 6889

Square root both sides.
Note that √(10x + y)² = 10x + y since x and y are positive and √6889 is 83

   10x + y = 83

Now use the other information to make another equation
The sum of its digits is 11 so

   x + y = 11

We have a system of equations as x and y must satisfy both equations for the question conditions to be met:

   10x + y = 83 ... (I)
   x + y = 11 ... (II)

This can be solved using elimination. Multiplying (II) by -1 to aim to get rid of y, we get

   10x + y = 83 ... (I)
   - x - y = -11 ... (II)

Adding the two equations, we get

   (10x - x) + (y - y) = (83 - 11)
which is
   9x = 72

Divide both sides by 9 to get

   x = 8

This is the tens digit.
Can use x + y = 11 ... (II) to figure out value of y, 

   x + y = 11 
   ⇒ y = 11 - x
   ⇒ y = 11 - 8
   ⇒ y = 3

x = 8 and y = 3
x is the tens-place digit
y is the ones-place digit

The number is 83

Answer 2

If a number multiplied by itself is $6889$, the number is $\sqrt{6889}=83$.

If we didn't know that, consider our two digit number whose square ends in 9.  Its ones digits must be 3 or 7, because those are the ones whose squares will end in 9. So we try $83^2$ and $47^2$, and 83 works.