Question

In a 2 digit number the digit at the unit place is equal to the square of the digit at a 10 place if the 5040 is added to the number digit and touch change find the number.​

Answer 1

Answer:

The answer is 39.

Step-by-step explanation:

Suppose that,

The two digit number is 10x+y

Then, the unit palace is y and tens palace is x

According to given question,

$y = {x}^{2}$

Then,

Again, according to given question,

$10x + y = 54 + 10y + x \\ = 10x - x + y - 10y + 54 = 0 \\ = 9x - 9y + 54 = 0 \\ = x - y + 6 = 0$

From equation (1) to,

$x - {x}^{2} + 6 = 0 \\ = - {x}^{2} + x + 6 = 0 \\ = {x}^{2} - x - 6 = 0 \\ = {x}^{2} - (3 - 2)x - 6 = 0 \\ = {x}^{2} - 3x + 2x - 6 = 0 \\ = x(x - 3) + 2(x - 3) = 0 \\ = (x - 3)(x + 2) = 0$

For,

$x - 3 = 0 \\ x = 3$

Put in (2) equation

Then,

$x - y + 6 = 0 \\ 3 - y + 6 = 0 \\ - y + 9 = 0 \\ y = 9$

Then,

The number is 10x+y

$= 10 \times 3 + 9 \\ = 39$