Solve. The sum of the squares of the digits of a positive two-digit number is 61, and the tens digit is 1 more than the units digit. Find the number.

Question

A) 45
B) 56
C) 65
D) 54

Answer 1

Let x = tens digit

     y = units digit

 

Then, x² + y² = 61 and x = y + 1

 

So, (y+1)² + y² = 61

 

      2y² + 2y + 1 = 61

 

      2y² + 2y - 60 = 0

 

        y² + y - 30 = 0

 

       (y+6)(y-5) = 0

 

       y = -6 or y = 5

 

Since y can't be negative, y = 5

 

So, y = 5 and x = 6

 

The number is 65.  

 

 

Answer 2

Looking at your answers, keep in mind that the tens digit has to be one more than the units(ones) digit.

 

A) This number, 45, has a tens digit of 4 and ones digit of 5. This can't be the answer because the tens is not one bigger.

 

B) This number, 56, has a tens digit of 5 and ones digit of 6. This can't be the answer.

 

C) and D) are possible answers

 

Do the squares of the digits sum to 61?

 

C) 5² + 6² = 25 + 36 = 61

 

This is your answer

 

Answer 3

A. 16 + 25 = 41 ≠ 61

 

Keep going.