1000 Followers Problem!

Write equation as $\frac{a'b-b'a}{b^2}=3 \frac{a}{b}$ This equation yields $a'b-b'a=3ab$ . Let $\displaystyle a=\prod_{i=1}^{k} p_i^{\alpha_i}$ , and $\displaystyle b=\prod_{j=1}^{l} q_j^{\beta_j}$ . We get $a \sum_{i=1}^{k} \frac{\alpha_i}{p_i} b - b \sum_{j=1}^{l} \frac{\beta_j}{q_j} a =3ab \ \ \ \ \Longrightarrow \ \ \ \ \sum_{i=1}^{k} \frac{\alpha_i}{p_i} - \sum_{j=1}^{l} \frac{\beta_j}{q_j} =3 \qquad (1)$ Multiply both sides of this equation by $\prod_{i=1}^{k} p_i \prod_{j=1}^{l} q_j$ to get $\prod_{j=1}^{l} q_j \sum_{i=1}^{k} \alpha_i \frac{\prod_{r=1}^{k} p_r }{p_i} - \prod_{i=1}^{k} p_i \sum_{j=1}^{l} \beta_j \frac{\prod_{s=1}^{l} q_s}{q_j} =3\prod_{i=1}^{k} p_i \prod_{j=1}^{l} q_j$ $a$ and $b$ share no common primes. Notice that $p_i$ divides all of the terms in the above equation except the $i$ -th term in the first summation thus $p_i \mid \alpha_i$ . Similarly $q_j$ divides all of the terms except $j$ -th term in the second summation, thus $q_j \mid \beta_j$ . Let $\alpha_i=A_i p_i$ and $\beta_j= B_j q_j$ , where $A_i$ and $B_j$ are non-negative integers.

It follows that $\displaystyle a=\prod_{i=1}^{k} p_i^{A_i p_i}$ , and $\displaystyle b=\prod_{j=1}^{l} q_j^{B_j q_j}$ . Thus none of $a$ or $b$ can contain a prime divisor greater than $3$ . Because otherwise they will be greater than $5^5$ , which exceeds 1000.

If $b=1$ then all of $B_j$ 's are zero and we have $n=a=2^{\alpha_1} 3^{\alpha_2}$ and from equation $(1)$ : $A_1+A_2=3$ . Hence possible solutions from here are $n=3^9, 2^2 \times 3^6, 2^4 \times 3^3, 2^6$ and two of them are less than 1000. $n= {\color{#D61F06} 2^4 \times 3^3}, {\color{#D61F06}2^6}$ .

For $1<b<a<1000$ , since $a$ and $b$ are co-prime, we must have $a=2^{\alpha_1}$ , $b=3^{\beta_1}$ or $a=3^{\alpha_1}$ , $b=2^{\beta_1}$ .

• For $a=2^{\alpha_1}$ and $b=3^{\beta_1}$ , from equation $(1)$ : $A_1=B_1+3$ . $B_1$ can take $1, 2$ only. $B_1=2$ gives $A_1=5$ and $a=2^{10}>1000$ . Thus $B_1=1$ and $A_1=4$ , which results in $n= \color{#D61F06} 2^8/3^3$ .

• For $a=3^{\alpha_1}$ and $b=2^{\beta_1}$ , from equation $(1)$ : $A_1=B_1+3$ . It gives $A_1 \ge 4$ and $a \ge 3^{12}>1000$ . Thus there is no solution in this case.

Therefore, sum of solutions is $2^4 \times 3^3+2^6+\dfrac{2^8}{3^3}=\dfrac{13648}{27}$ . And the answer is $\boxed{13675}$ .