Isn't $100000$ such a huge number (Part 3)?

Let $N = 10^5$ . Suppose $\lfloor \frac{x}{N} \rfloor = n$ ; this is equivalent to $x = Nn+r$ for some nonnegative $r < N$ . Then the given equation is equivalent to $\frac{x}{N-2} < n+1$ , which is equivalent to $Nn+r < (N-2)(n+1)$ , which is equivalent to $r+2n < N-2$ .

Each ordered pair $(r,n)$ (except for $(0,0)$ !) satisfying this inequality leads to a unique positive $x = Nn+r$ satisfying the equation. So we need to count these ordered pairs. Well, as $n$ ranges from $0$ to $N/2-2$ , $r$ ranges from $0$ to $N-2-2n$ . So the total number of pairs (remembering to subtract 1 to avoid $(0,0)$ ) is $\begin{aligned} -1 + \sum_{n=0}^{N/2-2} (N-2-2n) &= -1 + (N-2)(N/2-1)-(N/2-2)(N/2-1) \\ &= \frac{N(N-2)}4 - 1, \end{aligned}$ and plugging in $N = 100000$ gives our answer of $\fbox{2499949999}$ .

• When LHS equals 0, we have 99997 possible values of $x$ (0 is excluded).
• When LHS equals 1, we have 99996.
• When LHS equals 2, we have 99994.
• When LHS equals 3, we have 99992.
• It will continue until there is no more possible values of $x$ , and adding them together yields a sum of $2499949999$ (if you don't know how to add them together, level 5 is not the place for you :ppp) (no offense)

This is not a solution but a useful tool for such problem .

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I found generalization of such problem in a book without its proof. It is as follow - For any $m\geq 2$ the number of positive integers x such that $\bigg [ \frac {x}{m-1} \bigg ]=\bigg [ \frac {x}{m+1} \bigg ]$ is $\frac {m^{2}-4}{4}$ if m is even and $\frac {m^{2}-5}{4}$ , if m is odd .

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Here $[x]$ denotes greatest integer less than or equal to $x$ .

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Here m=99999 and m is odd so $\frac {99999^{2}-5}{4}$ = $\boxed {2499949999}$ .

The main thing is to see that for all numbers $x<99998$ , x is valid, but for $x = 99998$ , the LHS becomes 1 while RHS will still be 0. In the same way, this will continue until $x = 100000$ , when both LHS and RHS becomes 1.

And this will continue until $x = 99998*2$ and so on.

Hence, we can see that

number of solutions = $99997 + \sum _{ k=2 }^{ n }{ 99998(k) } -\quad 100000\left( k-1 \right)$

Now n for sigma operation can be found using the fact that we don't need negative values. So, n = 50000 and sigma operation further simplifies too.

number of solutions = $99997 + \sum _{ k=2 }^{ n }{ 99998 } -\quad 2k\quad$

= $99997 + (100000\times 49999)\quad -\quad 2(\frac { 50000\times 49999 }{ 2 } -1)$

= $\boxed{2499949999}$