+303 If $a$ and $b$ are positive integers for which $ab - 3a + 4b = 131$, what is the minimal possible value of $|a - b|$?
+1790 Ok, so we start off with the equation
\(ab -3a + 4b = 131 \)
I know this is random, but let'st ake the product of the coefficients on a,b.....add this product to both sides
So this gets us
\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119 \)
This eventually achieves the equation
\((a + 4) ( b - 3) = 119\)
Now, the factors of 119 are \(1, 7 , 17 , 119\)
Since we want the minimum, let's use 7 and 17. We have
\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)
Thus, our final answer is 3.
Thanks! :)
+1790 Ok, so we start off with the equation
\(ab -3a + 4b = 131 \)
I know this is random, but let'st ake the product of the coefficients on a,b.....add this product to both sides
So this gets us
\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119 \)
This eventually achieves the equation
\((a + 4) ( b - 3) = 119\)
Now, the factors of 119 are \(1, 7 , 17 , 119\)
Since we want the minimum, let's use 7 and 17. We have
\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)
Thus, our final answer is 3.
Thanks! :)
