Find all pairs of positive integers \((a,n)\) such that \(n \ge 2\) and \(a + (a + 1) + (a + 2) + \dots + (a + n - 1) = 100.\)
I would like a solid algebra solution not just guessing.
+983 Hi Guest!
We can rewrite the left side of the equation,
using the arithmetic series sum formula:
\(S_n=\frac{n(a_1+a_n)}{2}\)
We get:
\(100=\frac{n(2a+n-1)}{2}\)
\(200=n(2a+n-1)\)
Hence, n must be a factor of 200.
\( \begin{array}{c|c|c} n & 2a + n - 1 & a \\ \hline 2 & 100 & 99/2 \\ 4 & 50 & 47/2 \\ 5 & 40 & 18 \\ 8 & 25 & 9 \\ 10 & 20 & 11/2 \\ 20 & 10 & -9/2 \\ 25 & 8 & -8 \\ 40 & 5 & -17 \\ 50 & 4 & -45/2 \\ 100 & 2 & -97/2 \\ 200 & 1 & -99 \end{array} \ \)
Since a must be a positive integer, the solutions are \((a,n) = \boxed{(18,5)\text{ and }(9,8)}\).
I hope this helped,
Gavin
+983 Hi Guest!
We can rewrite the left side of the equation,
using the arithmetic series sum formula:
\(S_n=\frac{n(a_1+a_n)}{2}\)
We get:
\(100=\frac{n(2a+n-1)}{2}\)
\(200=n(2a+n-1)\)
Hence, n must be a factor of 200.
\( \begin{array}{c|c|c} n & 2a + n - 1 & a \\ \hline 2 & 100 & 99/2 \\ 4 & 50 & 47/2 \\ 5 & 40 & 18 \\ 8 & 25 & 9 \\ 10 & 20 & 11/2 \\ 20 & 10 & -9/2 \\ 25 & 8 & -8 \\ 40 & 5 & -17 \\ 50 & 4 & -45/2 \\ 100 & 2 & -97/2 \\ 200 & 1 & -99 \end{array} \ \)
Since a must be a positive integer, the solutions are \((a,n) = \boxed{(18,5)\text{ and }(9,8)}\).
I hope this helped,
Gavin
