Question

a two digit number is four times the sum of its digits and twice the product of its digits. find the number.

Answer 1

Let the digit at 10 place be x and the digit at ones place be y.

According to question,
10x+y=4(x+y)
10x+y=4x+4y
6x-3y=0
2x-y=0
y=2x -------eq1

10x+y=2xy -------eq2

putting the value of y in eq2
10x+2x=4x²
4x²=12x
x=3

putting the value of x in eq1
y=2×3=6

Therefore, the required digit is:
10x+y=10×3+6=36(Ans).

Answer 2

$\huge\underline\mathrm{SoLution:-}$

⭐ Let the digit at unit's place be 'y' and the digits at ten's place be 'x'. Then,

⚫ Number = 10x + y

According to given conditions, we have

$\longrightarrow$ 10x + y = 4(x + y)

$\longrightarrow$ 10x + y = 4x + 4y

$\longrightarrow$ 10x - 4x = 4y - y

$\longrightarrow$ 6x = 3y

Divide by '3' on both sides

2x = y -----1)

⭐ In the second condition,

$\longrightarrow$ 10x + y = 2xy ----2)

put the value of 'y' in equation 2)

$\longrightarrow$10x + 2x = 2x(2x)

$\longrightarrow$ 12x = 4x²

$\longrightarrow$ x= 3

put the value of 'x' in equation 1)

$\longrightarrow$ 2(3) = y

$\longrightarrow$ y = 6

⚫ On solving the equation 1) & 2), we get x= 3 and y = 6

$\implies$ Hence, number = 10x + y

➡ 10(3) + 6 = 36

$\huge\underline\mathrm{ThAnKs...}$