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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.

May 4, 2018

#1
+985
+4

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

May 4, 2018
edited by GYanggg  May 4, 2018

#1
+985
+4

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

GYanggg May 4, 2018
edited by GYanggg  May 4, 2018
#2
0

Thanks so much, I was looking desperatley for a way to do it without trial and error, but with two variables and one equation I guess that's unavoidable.

May 4, 2018