+219 If $a$ and $b$ are positive integers for which $ab - 3a + 4b = 131$, what is the minimal possible value of $|a - b|$?
+1926 Let's use a handy trick to solve this problem. We first have
\(ab -3a + 4b = 131 \)
Now, let's take the coefficients of a and b, which are -3 and 4. We multiply them and take the product and add it to both sides. We get
\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \)
Now, we simplify and factor the left side of the equation. We get that
\(ab - 3a + 4b - 12 = 119 \\ (a + 4) ( b - 3) = 119 \)
Now, let's focus on the factors of 119. We have
\(1 , 7 , 17, 119\)
Clearly, 7 and 17 will get us the minimal value, so plugging that in, we have
\(( 13 + 4) ( 10 - 3) → |a - b | = |13 - 10 | = 3 \)
so 3 is our final answer.
Thanks! :)
