Question

If we divide a two-digit number by the sum of its digits, we get $6$ as a quotient and $2$ as a remainder. Now if we divide it by the product of its digits, we get $5$ as a quotient and $2$ as a remainder. Find the number.

Answer 1

Let $n(a,b)$ be the number

$n(a,b)=10a+b$

Given that $n(a,b)=6(a+b)+2$

$10a+b=6a+6b+2$

$4a=5b+2$

$n(a,b)=5ab+2$

$16a+b=5ab+2$

$10a+\frac{4a-2}{5}=a(4a-2)+2$

$50a+4a-2=20a^2-10a+10$

$20a^2-64a+12=0$

$5a^2-16a+3=0$

$(5a-1)(a-3)=0$

$a=3,a\ne \frac{1}{5}$

$\therefore$ $32$ is the required number.