p1. There are two glasses, glass $A$ contains $5$ red balls, and glass $B$ contains $4$ red balls and one white ball. One glass is chosen at random and then one ball is drawn at random from the glass. This is done repeatedly until one of the glasses is empty. Determine the probability that the white ball is not drawn.

p2. For any real number $x$, define as $[x]$ the largest integer less than or equal to $x$. Find the number of natural numbers $n  \le 1,000,000$ such that $$\sqrt{n}-[\sqrt{n}] <\frac{1}{2013}$$

p3. A natural number $n$ is said to be  [i]valid[/i] if $1^n + 2^n + 3^n +... + m^n$ is divisible by $1 + 2 + 3 +...+ m$ for every natural number $m$.
a)  Show that $2013$ is valid.
b)  Prove that there are infinitely many invalid numbers.

[url=https://artofproblemsolving.com/community/c4h2685414p23296970]p4.[/url] Prove that for all positive real numbers $a, b, c$ where $a + b + c\le 6$ holds
$$\frac{a+2}{a(a+4)}+\frac{b+2}{b(b+4)}++\frac{c+2}{c(c+4)} \ge 1$$

[url=https://artofproblemsolving.com/community/c6h2371585p19388892]p5.[/url] Given an acute triangle $ABC$. The longest line of altitude is the one from vertex $A$ perpendicular to $BC$, and it's length is equal to the length of the median of vertex $B$. Prove that $\angle ABC \le 60^o$
To solve the problem, we need to find the number of natural numbers \( n \leq 1,000,000 \) such that 
\[ \sqrt{n} - \lfloor \sqrt{n} \rfloor < \frac{1}{2013}. \]

1. **Case 1: \( n \) is a perfect square**

   If \( n \) is a perfect square, then \( n = k^2 \) for some integer \( k \). In this case, 
   \[ \sqrt{n} = k \]
   and 
   \[ \lfloor \sqrt{n} \rfloor = k. \]
   Therefore,
   \[ \sqrt{n} - \lfloor \sqrt{n} \rfloor = k - k = 0 < \frac{1}{2013}. \]
   This inequality is always satisfied for perfect squares.

2. **Case 2: \( n \) is not a perfect square**

   Let \( n = a^2 + b \) where \( 1 \leq b \leq 2a \) and \( a, b \in \mathbb{N} \). In this case,
   \[ \lfloor \sqrt{n} \rfloor = a. \]
   We need to analyze the inequality:
   \[ \sqrt{n} - \lfloor \sqrt{n} \rfloor = \sqrt{a^2 + b} - a < \frac{1}{2013}. \]

   To understand this, we approximate:
   \[ \sqrt{a^2 + b} \approx a + \frac{b}{2a}. \]
   Therefore,
   \[ \sqrt{a^2 + b} - a \approx \frac{b}{2a}. \]

   For the inequality to hold,
   \[ \frac{b}{2a} < \frac{1}{2013}. \]
   This implies:
   \[ b < \frac{2a}{2013}. \]

   Since \( b \) is an integer and \( 1 \leq b \leq 2a \), the above inequality is generally not satisfied for non-perfect squares because \( b \) would need to be very small compared to \( a \).

3. **Counting the valid \( n \)**

   From the above analysis, the inequality \( \sqrt{n} - \lfloor \sqrt{n} \rfloor < \frac{1}{2013} \) is satisfied if and only if \( n \) is a perfect square. 

   The number of perfect squares \( n \leq 1,000,000 \) is the number of integers \( k \) such that \( k^2 \leq 1,000,000 \). This is equivalent to:
   \[ k \leq \sqrt{1,000,000} = 1000. \]

   Therefore, there are \( 1000 \) perfect squares less than or equal to \( 1,000,000 \).

The final answer is \( \boxed{ 1000 } \).