I've Just Solved 1236 Problems on Brilliant

Let $A_P$ be the set of all positive integers $P$ with $2n>0$ digits and the leftmost digit is non-zero such that the sum of digits of $P$ is equal to its Left-Half and the product of digits of $P$ is equal to its Right-Half .

We also use $S_P$ to denote the sum of digits of $P$ and $G_P$ to denote the product of digits of $P$ .

We consider the decimal expansion of $P=d_{2n-1}\times10^{2n-1}+d_{2n-2}\times10^{2n-2}+ \ldots{ } \ldots{ } \ldots{ }+d_{n}\times10^{n}+\ldots{ } \ldots{ } \ldots{ }+d_1\times10+d_0$ , with $d_{2n-1}\neq0$ .

Then $L=d_{2n-1}\times10^{n-1}+d_{2n-2}\times10^{n-2}+ \ldots{ } \ldots{ } \ldots{ }+d_{n}$ and $R=d_{n-1}\times10^{n-1}+d_{n-2}\times10^{n-2}+ \ldots{ } \ldots{ } \ldots{ }+d_{0}$ .

First let's prove such $P$ do not exist for $n>2$ . To prove this, we'll show that $L>S_P$ for $n\geq 3$ .

As $d_{2n-1}\neq0$ , we have $L\geq 10^{n-1}$ . We also have $18n\geq S_P$ . So, it suffices to show that $10^{n-1}>18n$ for $n\geq3\ldots{ } \ldots{ } \ldots{ } (1)$ . We apply the Principle of Mathematical Induction on $n$ .

The inequality $(1)$ holds for $n=3$ ; $10^{3-1}=10^2=100>54=18\times3$ .

Assume that the inequality $(1)$ holds for $n=m$ ; that is, $10^{m-1}>18m$ .

Now, $10^{m-1}>18m$

$\implies 10^{m-1}+18>18m+18$ (Adding $18$ to both sides)

$\implies 10^{m-1}+9\times 10^{m-1} >10^{m-1}+18 > 18m+18$ (By inductive hypothesis)

$\implies 10\times10^{m-1}>18(m+1)$

$\implies 10^{m}>18(m+1)$

$\implies 10^{(m+1)-1}>18(m+1)$ , that is, the inequality $(1)$ holds for $n=m+1$ .

So, by PMI , $10^{n-1}>18n$ for $n\geq3$ . That means, such $P$ don't exist for $n>2$ .

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Let's see if there are such $P$ for $n=1$ .

For $n=1$ , the decimal expansion of $P=d_1\times 10+d_0$ . Then, $L=d_1$ , $R=d_0$ .

$P\in A_P$ only if $S_P=d_1+d_0=L=d_1 \implies d_1+d_0=d_1 \implies d_0=0$ .

So, $d_0=0$ is a necessary condition for $P\in A_P$ .

Moreover, it turns out that $d_0=0$ is also a sufficient condition for $P\in A_P$ . Because, if $d_0=0$ , then $S_P=d_1+d_0=d_1=L$ and $G_P=d_1\times d_0=0=d_0=R$ .

That means, $P=d_1\times 10+d_0 \in A_P \iff d_0=0$ .

So, $\boxed{10,20,30,40,50,60,70,80, 90 \in A_P}$ .

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Now let's see if there are such $P$ for $n=2$ .

For $n=2$ , the decimal expansion of $P=d_3\times 10^3+d_2\times10^2+d_1\times 10+d_0$ . Then $L=d_3\times10+d_2$ and $R=d_1\times10+d_0$ .

Now, $L=S_P$

$\iff d_3\times10+d_2=d_3+d_2+d_1+d_0$

$\iff 9\times d_3=d_1+d_0$

As $d_1+d_0\leq18$ and $d_3\neq0$ , there are two possibilities for $d_3$ , namely, $1$ and $2$ .

When $d_3=2$ , we have only one option for $R=d_1\times10+d_0$ , namely $R=99$ . Then $P=2a99$ for some decimal digit $a$ . But $P=2a99$ satisfies the $G_P=R$ condition for NO $a$ . So, $P=2a99 \not \in A_P$ .

When $d_3=1$ , the options we have for $R=d_1\times10+d_0$ are: $09, 90, 18, 81, 27, 72, 36, 63, 45$ and $54$ . Applying the same logic on each $R$ as we did in the case when $d_3=2$ , we'll able to find only one $P$ that belongs to $A_P$ , namely, $1236$ .

Hance, $\boxed{A_P=\{10,20,30,40,50,60,70,80, 90,1236\}}$ .

So, the answer is $10+20+30+40+50+60+70+80+90+1236=\boxed{1686}$ .