The sum of both digits, of either of two two-digit numbers, in whatever order the digits are written, is 9. T

Question

The sum of both digits, of either of two two-digit numbers, in whatever order the 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 _______?
a. 36,63b. 81, 18c. 27, 72d. 45, 54e. none

Please show the solution...

Answer 1

We know that if x represents the first (10s) digit, and y represents the ones digit, then

n = 10x + y

We're also given that the sum of the digits is 9.

x + y = 9

And this information as well:

x^2 - xy + y^2 = 21

If we add 3xy to both sides, we get

x^2 + 2xy + y^2 = 21 + 3xy

We can factor the left hand side,

(x + y)^2 = 21 + 3xy

Note that, as stated earier, (x + y) = 9. Therefore, we can totally replace (x + y) with 9 in this case.

81 = 21 + 3xy

Simplifying,

60 = 3xy

20 = xy

This means the product of x and y is 20. All you have to do now is glance at your choices.
It's not (a), because 6 x 3 = 18
It's not (b), because 8 x 1 = 8
It's not (c), because 7 x 2 = 14
The answer is (d), because 4 x 5 = 20.

Answer 2

let x= digit 1 of 1st number
y = digit 2 1st number
a = digit 1 2nd number
b = digit 2 2nd number

x2 - x*y + y2 = 21
x+y = 9
to get the perfect square of x+y squared we add 3xy to both sides of the equation thus we have
(x+y)2 = 21+3xy
simplifying...
81-21=3xy
xy=20
looking for the factors of 20 which has a sum of 9
20 = 20*1 20+1 =21 wrong answer
20 = 10*2 10+2 =12 wrong answer
20 = 5*4 5+4 = 9 well well dont look now but there is your answer letter d. 54 and 45

Answer 3

There is yet another, more algebraic, way of doing it, which does not involve trial and error in the last step.

If the sum of the digits is 9, then if one digit is z, the other digit is (9-z).

Now, you need the square of one digit, minus their product, plus the square of the other digit. So you have:
Square of one digit: z^2
Product: z(9-z) = 9z - z^2
Square of other digit: (9-z)^2 = 81 - 18z + z^2

The sum of the squares, minus the product, is 21.

So:
z^2 + [81 - 18z + z^2] - [9z - z^2] = 21
z^2 + 81 - 18z + z^2 -9z + z^2 - 21 = 0
3z^2 - 27z + 60 = 0
(3z - 12)(z - 5) = 0

So z = 4 or z = 5, which of course gives the other digit as 5 or 4, respectively.
So you numbers are 45 and 54.

Answer 4

I guess I would like to know where you find 3xy out of thin air to solve the equation.

Answer 5

ans is (d)