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The product of two consecutive integers, n and n + 1 is 42. What is the positive integer that satisfies the situation?

 Jul 2, 2019
 #1
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Six

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 Jul 2, 2019
 #2
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Let's rewrite your problem in mathematical terms:

\((n)(n+1)=42\)

We can multiply the left side to get:

\(n^2+n=42\)

We can move 42 to the left side to get:

\(n^2+n-42=0\)

From here, we just solve for \(n\).

\((n+7)(n-6)=0\)

In order for this to be true, we need to either satisfy \(n-6=0\), meaning that \(n=6\), or \(n+7=0\), meaning that \(n=-7\). We need to find the positive solution for n, so we pick \(n=6\) . This does indeed work, because 6*7=42, so \(\boxed{n=6}\)

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 Jul 2, 2019

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