Question

A two-digit number is thrice as large as the sum of its digits, and the square of that sum is equal to the trippled required number. Find the number.

Answer 1

Let the two digit number be $=10x+y$

Given,

$(x+y)3=10x+y$(equation $1$)

and $(x+y{)}^2=(10x+y)3$(equation $2$)

From equation $1$,

$3x+3y=10x+y$

$\Rightarrow 7x-2y=0$(equation$3$)

$\Rightarrow x=\frac{2y}{7}$

Substituting $x=\frac{2y}{7}$ in equation $2$,

$(x+y{)}^2=(10x+y)3$

$\Rightarrow {(\frac{2y}{7}+4)}^2=(10\left(\frac{2y}{7}\right)+y)3$

$\Rightarrow {\left(\frac{9y}{7}\right)}^2=(\frac{20y}{7}+y)3$

$\Rightarrow \frac{81y^2}{49}=\frac{27y}{7}\times 3$

$\Rightarrow y=7$

So, $x=2$

The required number is $27$