# Math Help - the House Problem

1. ## the House Problem

Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
(hint: N is a 4 digits number)

2. Originally Posted by john_n82
Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
(hint: N is a 4 digits number)

Note, $1+2+...+(N-1) = \tfrac{1}{2}N(N-1) \text{ and }(N+1)+(N+2)+...+M =$ $\tfrac{1}{2}(M-N)(M-N+1) + N(M-N)$

Therefore, $N^2 = \tfrac{1}{2}M(M+1)$, therefore $M^2 + M - 2N^2 = 0$.
We need the discriminant to be a square, so we need $1 + 8N^2 = x^2$.
This is a Pellian equation.

3. Originally Posted by ThePerfectHacker
Note, $1+2+...+(N-1) = \tfrac{1}{2}N(N-1) \text{ and }(N+1)+(N+2)+...+M =$ $\tfrac{1}{2}(M-N)(M-N+1) + N(M-N)$

Therefore, $N^2 = \tfrac{1}{2}M(M+1)$, therefore $M^2 + M - 2N^2 = 0$.
We need the discriminant to be a square, so we need $1 + 8N^2 = x^2$.
This is a Pellian equation.
I found M = 8, so N = 6.

4. Originally Posted by john_n82
Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
(hint: N is a 4 digits number)
i think you mean N^2 is a four-digit number

in this case M = 49, N = 35

5. i found 2 more values for M

M = 288, N = 204
M 9800, N = 6930

What is the exact answer for this problem?

6. Originally Posted by john_n82
i found 2 more values for M

M = 288, N = 204
M 9800, N = 6930

What is the exact answer for this problem?
There are many answers to this problem.
You are told that you need N to be a 4 digit number so N=6930