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# Algebra

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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|\$?

Jul 25, 2024

#1
+1655
+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! :)

Jul 25, 2024
edited by NotThatSmart  Jul 25, 2024

#1
+1655
+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
edited by NotThatSmart  Jul 25, 2024