If $a$ and $b$ are positive integers for which $ab - 3a + 4b = 131$, what is the minimal possible value of $|a - b|$?
We can use a really simple and cheeky trick to solve this problem.
Taking the product of the coeffiicents of a and b and add this to both sides, we get
\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119 \)
Factoring the left side of the equation, we get
\((a + 4) ( b - 3) = 119\)
Now, we take the two factros with the smallest margin, 7 and 17. We have
\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)
So 3 is our answer.
Thanks! :)
We can use a really simple and cheeky trick to solve this problem.
Taking the product of the coeffiicents of a and b and add this to both sides, we get
\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119 \)
Factoring the left side of the equation, we get
\((a + 4) ( b - 3) = 119\)
Now, we take the two factros with the smallest margin, 7 and 17. We have
\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)
So 3 is our answer.
Thanks! :)