Question

A two digit number is $5$ times the sum of its digits and is also equal to $5$ more than twice the product of its digits. Find the number.

Answer 1

Let the number be $10x+y$

According to question, we have,

$10x+y=5(x+y)$

$=>10x+y=5x+5y$

$=>5x=4y$

Also, $10x+y-2xy=5$

$=>10x+\frac{5x}{4}-2x\times \frac{5x}{4}=5$

$=>10x+\frac{5x}{4}-\frac{10x^2}{4}=5$

$=>40x+5x-10x^2=20$

$=>10x^2-45x+20=0$

$=>2x^2-9x+4=0$

$=>2x^2-8x-x+4=0$

$=>2x(x-4)-1(x-4)=0$

$=>(2x-1)(x-4)=0$

$=>x=4$

$y=\frac{5x}{4}=\frac{5\times 4}{4}=5$

Required number is $10\times 4+5=45$