1989, Russian Mathematical Olympiad

Note that $n = d_jd_{13-j}$ for any $1 \le j \le 12$ . Since $d_{d_4-1} = d_8(d_1+d_2+d_4)$ , we must have $8 < d_4-1 \le 12$ , and hence $9 < d_4 \le 13$ .

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If $d_4=10$ then $d_2=2$ and $d_3=5$ . Moreover $d_5n \; = \; d_5d_4d_9 \; = \; d_5d_4(d_1+d_2+d_4)d_8 \; = \; d_4(d_1+d_2+d_4)n$ so that $d_5 = d_4(d_1+d_2+d_4) = 130$ . But then $13$ is a divisor of $n$ and $d_4 < 13 < d_5$ . This case is impossible.

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If $d_4=11$ then $d_5n \; = \; d_5d_3d_{10} \; = \; d_5d_3(d_1+d_2+d_4)d_8 \; = \; d_3(d_1+d_2+d_4)n$ so that $d_5 = d_3(d_1+d_2+d_4)$ .

• If $d_2=2$ then $d_1+d_2+d_4=14$ divides $n$ , so that $d_3=7$ , and so $d_5=98$ . But then $49$ divides $n$ and $d_4 < 49 < d_5$ .
• If $d_2=3$ then $d_1+d_2+d_4=15$ divides $n$ , so that $d_3=5$ , and so $d_5=75$ . But then $25$ divides $n$ and $d_4 < 25 <d_5$ .
• If $d_2=5$ then $d_3=7$ and $d_5=119$ . But then $17$ divides $n$ and $d_4 < 17 < d_5$ .

None of these cases are possible

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If $d_4=12$ then $2,3,4,6$ are all divisors of $n$ , which is impossible.

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Thus we deduce that $d_4=13$ , and so $n = (d_1+d_2+d_4)d_8$ , so that $d_5 = d_1 + d_2 + d_4$ ..

• If $d_2=2$ then $d_1+d_2+d_4=16$ divides $n$ , so that $4,8$ are also divisors of $n$ . This case is impossible.
• If $d_2=3$ then $d_1+d_2+d_4 =17$ divides $n$ .
• If $d_2=5$ then $d_1+d_2+d_4=19$ divides $n$ .
• If $d_2=7$ then $d_1+d_2+d_4=21$ divides $n$ , so that $3$ is a divisor of $n$ , which is impossible.
• If $d_2=11$ then $d_3=12$ , which is impossible.

After running through all these cases, there are only six possible values of $n$ , namely $3^2\times13\times17$ , $3\times13^2\times17$ , $3\times13\times17^2$ , $5^2\times13\times19$ , $5\times13^2\times19$ , $5\times13\times19^2$ . Checking these cases shows that the correct solution is $3^2 \times 13 \times 17 = \boxed{1989}$ .