If we divide a 2-digit positive integer by the sum of its digits, we get the quotient and remainder of 4 and 3, respectively.

If we divide the same 2-digit positive by the product of its digits, we get quotient and remainder of 3 and 5, respectively.

If this 2-digit integer can be written as $\overline{AB}$ , find $A+B$ .

The answer is 5.

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Let the number be $\overline{ab} = 10a+b$

Given that:

$\dfrac{10a+b}{a+b} = 4+\dfrac{3}{a+b}\\ 10a+b= 4(a+b) + 3\\ 10a+b=4a+4b+3\\ 3b = 6a-3$

$b=2a-1 \implies$ Eq.(1)

and

$\dfrac{10a+b}{ab} = 3+\dfrac{5}{ab}$

$10a+b= 3ab+5\implies$ Eq.(2)

Substitute Eq.(1) into Eq.(2):

$10a +(2a-1) = 3a(2a-1) +5\\ 12a-1 = 6a^2 - 3a + 5\\ 6a^2 - 15a + 6 = 0\\ 2a^2 - 5a + 2=0\\ (2a-1)(a-2) =0\\ a=\dfrac{1}{2},\;2$

Since $a$ represents a digit, it must be a positive integer between $1$ and $9$ . Therefore,

$a=2\\ b=2(2)-1 = 3\\ a+b = 2+3 = \boxed{5}$