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If $a$ and $b$ are positive integers for which $ab - 3a + 4b = 131$, what is the minimal possible value of $|a - b|$?

 Jun 9, 2024
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We can use Simon's Favourite Factoring Trick to solve this problem.

\(ab - 3a + 4b = 131\)

\(a(b-3)+4b=131\)

\(a(b-3) + 4b - 12 = 131 - 12\)

\(a(b-3)+4(b-3)=119\)

\((a+4)(b-3) = 119\)

Now, the only 4 pairs of numbers that multiply to 119 are (1, 119), (7, 17), (17, 7), or (119, 1). This means that the only possibilities for a and b are (-3, 122), (3, 21), (13, 10), and (115, 4). The absolute difference for each of these numbers is 125, 18, 3, and 111 respectively. The smallest absolute difference out of all of these is 3

 Jun 9, 2024

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