Christmas Streak 06/88: Sum Of The Divisors!

Firstly, let's prove a lemma.

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Lemma: $S(xy)\le S(x)S(y),$ and the equality holds when $x$ and $y$ are coprime.

Since it is clear that the equality holds when $x$ and $y$ are coprime, all we need to prove is $S(xy)<S(x)S(y)$ when $x$ and $y$ have common prime factors of $p_1,~p_2,~\cdots,~p_n.$

Let $x={p_1}^{a_1}\cdots{p_n}^{a_n}X$ and $y={p_1}^{b_1}\cdots{p_n}^{b_n}Y,$ where $\text{gcd}(X,~Y)=1,$ and you see that

$S(x)S(y)=(1+p_1+{p_1}^2+\cdots+{p_1}^{a_1})\cdots(1+p_n+{p_n}^2+\cdots+{p_n}^{a_n})S(X)\times(1+p_1+{p_1}^2+\cdots+{p_1}^{b_1})\cdots(1+p_n+{p_n}^2+\cdots+{p_n}^{b_n})S(Y);$

$S(xy)=(1+p_1+{p_1}^2+\cdots+{p_1}^{a_1+b_1})\cdots(1+p_n+{p_n}^2+\cdots+{p_n}^{a_n+b_n})S(X)S(Y).$

Then, all we need to prove is $(1+p+p^2+\cdots+p^a)(1+p+p^2+\cdots+p^b)>(1+p+p^2+\cdots+p^{a+b}).$

Are we sure that $p(p^a-1)(p^b-1)>0$ ? Good, let's start.

$\begin{aligned} p^{a+b+1}-p^{a+1}-p^{b+1}+p & > 0 \\\\ p^{a+b+2}-p^{a+1}-p^{b+1}+1 &> p^{a+b+2}-p^{a+b+1}-p+1 \\\\ (p^{a+1}-1)(p^{b+1}-1) &> (p^{a+b+1}-1)(p-1) \\\\ \frac{p^{a+1}-1}{p-1}\cdot\frac{p^{b+1}-1}{p-1} &> \dfrac{p^{a+b+1}-1}{p-1} \\\\ (1+p+p^2+\cdots+p^a)(1+p+p^2+\cdots+p^b) &>(1+p+p^2+\cdots+p^{a+b}) \end{aligned}$

And therefore the lemma is proved. $\square$

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$S(6x)\le S(6)S(x)=12S(x),$ and therefore, $S(6x)\ge12S(x)$ is only satisfied when the equality holds, which means $x$ and $6$ are coprime.

Since three-digit numbers range from $100$ to $999,$ there are $900$ of them.

And from that there are $2$ numbers that are relatively prime to $6$ among $1,~2,~3,~4,~5,~6,$

$\dfrac{1}{3}$ of the numbers are relatively prime to $6.$

$\therefore~900\times\dfrac{1}{3}=\boxed{300}.$