Question

If we divide a two-digit number by the sum of its digits, we get $4$ as a quotient and $3$ as a remainder. Now if we divide that two-digit number by the product of its digits, we get $3$ as a quotient and $$5 as a remainder and the two-digit number.

Answer 1

Let $n(a,b)$ is two digit number.

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

Given $\frac{number}{\text{sum of its digits}}$ will give $4$ as quotient and $3$ as remqinder.

$n(a,b)=4(a+b)+3$

$10a+b=4a+4b+3$

$6a-3b=3\Rightarrow 2a=1+b$

$n(a,b)=3\left(ab\right)+5$

$10a+b=3ab+5$

$10a+2a-1=3a(2a-1)+5$

$12a-1=6a^2-3a+5$

$6a^2-15a+6=0$

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

$(2a-1)(a-2)=0$

$a=2(\because a\ne \frac{1}{2})$

$\therefore b=2a-1=3$

$23$ is required number.