North Macedonia

We will use: Lemma 1. The number of solutions of the equation $x_1+x_2+\cdots +x_k=n$ 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 $x_1+x_2+\cdots +x_k=n$ 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 $$x_1+x_2+\cdots +x_k=n\text{\hspace{1em}(1)}$$ 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 $$y_1+y_2+\cdots +y_k=n+k-1,\text{\hspace{1em}(2)}$$ 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, $a_{5n}=a_{325}=52000$.