+41 The product of two consecutive integers, n and n + 1 is 42. What is the positive integer that satisfies the situation?
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 \(\boxed{n=6}\)
