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

onyuIee Jul 25, 2024

#1**+1 **

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! :)

NotThatSmart Jul 25, 2024

#1**+1 **

Best Answer

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! :)

NotThatSmart Jul 25, 2024