A and B are positive integers. If A + B + AB | vedclass

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(C) Given the equation: $A + B + AB = 65$.Add $1$ to both sides to factorize the expression:$A + B + AB + 1 = 65 + 1$$A(1 + B) + 1(1 + B) = 66$$(A + 1

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$5$

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(C) Given the equation: $A + B + AB = 65$.Add $1$ to both sides to factorize the expression:$A + B + AB + 1 = 65 + 1$$A(1 + B) + 1(1 + B) = 66$$(A + 1)(B + 1) = 66$Since $A$ and $B$ are positive integers,$(A + 1)$ and $(B + 1)$ must be factors of $66$ greater than $1$.The pairs of factors of $66$ are $(1, 66), (2, 33), (3, 22), (6, 11)$.Since $A, B \leq 15$,then $(A + 1) \leq 16$ and $(B + 1) \leq 16$.The only pair satisfying this condition is $(6, 11)$.Thus,$A + 1 = 6$ and $B + 1 = 11$ (or vice versa).$A = 5$ and $B = 10$.The difference between $A$ and $B$ is $|10 - 5| = 5$.

$A$ and $B$ are positive integers. If $A + B + AB = 65$,then what is the difference between $A$ and $B$ (given $A, B \leq 15$)?

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