North Macedonia competition problems

We will use:
Lemma 1. The number of solutions of the equation $x_1 + x_2 + \dots + x_k = n$ in the set of natural numbers is $C_{n-1}^{k-1}$.
Proof. Consider $n$ ones arranged in a row, and an arrangement of $k-1$ signs "+" between $k-1$ pairs of ones.
This arrangement of the "+" signs can be done in $C_{n-1}^{k-1}$ ways while every such possibility arranges the ones in $k$ non-empty subsets.
Let the number of ones in every one of the subsets be $m_1, m_2, \dots, m_k$ (clearly $m_i > 0$). Then $x_i = m_i, i = 1, 2, \dots, k$ is a solution of the first equation and every solution can be obtained in this way.
This ends the proof of the lemma.
Lemma 2. The number of solutions of the equation $x_1 + x_2 + \dots + x_k = n$ in the set of non-negative integers $(N_0)$ is $C_{n+k-1}^{k-1}$ i.e. $C_{n+k-1}^n$.
Proof. The proof is obtained directly from Lemma 1 by substituting $x_i = y_i - 1$, $i = 1, 2, ..., k$.
Lemma 3. The number of solutions of the equation
$$x_1 + x_2 + \dots + x_k = n \tag{1}$$
in the set of non-negative integers $(N_0)$, where $x_1 \ge 1$ is $C_{n+k-2}^{k-1}$ i.e. $C_{n+k-2}^{n-1}$.
Proof. We make substitutions $x_1 = y_1$, $x_i = y_i - 1$ for $i = 2, ..., k$; $x_1 = y_1$, $x_i = y_i - 1$ for $i = 2, ..., k$
Then the equation (1) is equivalent to the equation
$$y_1 + y_2 + \dots + y_k = n + k - 1, \tag{2}$$
where clearly for $i = 1, 2, ..., k$ it holds that $y_i = x_i + 1 \ge 1$ i.e. the numbers $y_i$ are natural numbers for every $i = 1, 2, ..., k$.
To every solution $(x_1, x_2, ..., x_k)$ of equation (1) corresponds a unique solution $(y_1, y_2, ..., y_k)$ of the equation (2) and vice-versa. Therefore we get that the number of solutions of equation (1) in the set of non-negative whole numbers under the condition $x_1 \ge 1$ is equal to the number of solutions of the equation (2) in the set of natural numbers which according to Lemma 1 is equal to $C_{n+k-2}^{k-1}$ i.e. $C_{n+k-2}^{n-1}$.
Now according to Lemma 3 for $n=7$, the number of “lucky” numbers having $k$ digits is
$$p(k) = C_{k+5}^{6}.$$
So we have
$p(1) = C_6^6 = 1$, $p(2) = C_7^6 = 7$ and $p(3) = C_8^6 = 28$.
The number of four-digit “lucky” numbers in the form $abc$ is actually equal to the number of solutions of the equation $a+b+c=6$ in the set of non-negative whole numbers and according to Lemma 2 that number is equal to $C_{6+3-1}^2 = C_8^2 = 28$. Now, $2005$ is the least 4-digit “lucky” number of the form $2abc$. Therefore $2005$ is the $1+7+28+28+1=65$-th in order “lucky” number.
So $a_{65} = 2005$ i.e. $n=65$ which implies $5n=325$.
Then we have $p(4) = C_9^6 = 84$, $p(5) = C_{10}^6 = 210$ which implies that
$$p(1) + p(2) + p(3) + p(4) + p(5) = 1 + 7 + 28 + 84 + 210 = 330.$$
So there is a total of $330$ “lucky” numbers having at most $5$ digits.
The least six “lucky” numbers arranged in decreasing order are $70000$, $61000$, $60100$, $60010$, $60001$ and $52000$.
Therefore, $a_{5n} = a_{325} = 52000$.