Question

in a 2 digit natural number the digit at the tens place is equal to square of the digit at the units place if 18 is subtracted from the number the digits get interchange .find the number

Answer 1

Answer:

x = 4, y = 2

Step-by-step explanation:

let the units be x, y

let the initial number be 10x+y = 10y²+y

since, x = y² (given)

ATQ,

10y²+y - 18 = 10y+y²_____(i)

=> 9y²-9y=18

=>y²-y= 2

=>y²-y-2=0

=>(y-1)(y+2)=0. {after factorizing}

hence, y = 1 or -2

After putting y = 1 in eq(i) we dont get required solution

hence y = -2 is the right answer

since, x= y²

x = (-2)² = 4

Edit: I forgot to find the numbers sorry!

its, 10(4)+(-2) = 38 is the number

Answer 2

Answer:

After calculation, the number is 42.

Step-by-step explanation:

We are given a $2$ digit natural number and the digit at the ten's place is equal to the square of the digit at the unit's place and if $18$ is subtracted from the number the digits get interchanged.

Let the digit at the unit's place $=x$

The digit at ten's place $=y$

The equation for the number will be $=10 y+x$

$y=x^2$

$10 y+x-18=10 x+y\\9y-9x=18\\x=\frac{18-9y}{-9} \\x=y-2$

Putting $x=y-2$ in the equation $y=x^2$:

$y=(y-2)^2\\y=y^2+4-4y\\y^2-5y+4=0\\y^2-4y-y+ 4=0\\y(y-4)-1(y-4)\\(y-1)(y-4)=0\\y=1,4$

Putting $y=1,4$ in the equation $x=y-2$:

$x=y-2\\x=1-2,x=4-2\\x=-1,2$

We will consider $x=2,y=4$, because $-1$ is a negative digit.

The equation for number: $10y+x$

$=10y+x\\=10(4)+2\\=42$

Thus, the number is 42.