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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 27, 2024

Best Answer 

 #1
avatar+1230 
+1

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! :)

 Jun 27, 2024
 #1
avatar+1230 
+1
Best Answer

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! :)

NotThatSmart Jun 27, 2024

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