The sum of both digits, of either of two-digit numbers, in whatever order thw digits are written, is 9. The square of either of the digits of either number, minus the product of both digits,

Question

The sum of both digits, of either of two-digit numbers, in whatever order thw digits are written, is 9. The square of either of the digits of either number, minus the product of both digits, plus the square of the other digit is the number 21. The numbers are?

Answer 1

The sum of both digits, of either of two-digit numbers, in whatever order thw digits are written, is 9. The square of either of the digits of either number, minus the product of both digits, plus the square of the other digit is the number 21. The numbers are?

Think of 2 numbers, the sum of which is 9:

1 and 8

2 and 7

3 and 6

4 and 5

Then think of the squares of each of those numbers

1^2=1

2^2=4

3^2=9

4^2=16

5^2=25

6^2=36

7^2=49

8^2=64

Then think of the product of both digits:

1*8 = 8

2*7 = 14

3*6 = 18

4*5 = 20

Then work backwards: (I happened to start at the bottom and found the answer right away)

21-25+20=16 so the numbers are 4 and 5.

But do the same process for the other sets of numbers and you will discover that it doesn't work:

For example: 21-36+18 does not equal 9

Answer 2

This can also be solved algebraically. Let the 2 digits be a and b, going left to right. The number itself is then

N = 10a + b. And it's given that a + b = 9, and a² - ab + b² = 21. Substitute for b = 9 - a, combine all terms of a² and a, and you get

3a² - 27a + 60 = 0. Divide by 3

a² - 9a + 20 = 0. Factor

(a - 5)(a - 4) = 0. So the roots are a = 5 and a = 4.

So the numbers N themselves are 54 and 45. Note the equations with only a and b are symmetric, so the digits are interchangeable. (if a = 5, b = 4 and vice versa)