1989, Russian Mathematical Olympiad

Find the positive integers n n with exactly 12 12 divisors 1 = d 1 < d 2 < < d 12 = n 1=d_1<d_2<\cdots<d_{12}=n such that the divisor with index d 4 1 d_4-1 ( that is , d d 4 1 d_{d_4-1} ) is ( d 1 + d 2 + d 4 ) d 8 (d_1+d_2+d_4)d_8 .


The answer is 1989.

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1 solution

Mark Hennings
Aug 21, 2019

Note that n = d j d 13 j n = d_jd_{13-j} for any 1 j 12 1 \le j \le 12 . Since d d 4 1 = d 8 ( d 1 + d 2 + d 4 ) d_{d_4-1} = d_8(d_1+d_2+d_4) , we must have 8 < d 4 1 12 8 < d_4-1 \le 12 , and hence 9 < d 4 13 9 < d_4 \le 13 .


If d 4 = 10 d_4=10 then d 2 = 2 d_2=2 and d 3 = 5 d_3=5 . Moreover d 5 n = d 5 d 4 d 9 = d 5 d 4 ( d 1 + d 2 + d 4 ) d 8 = d 4 ( d 1 + d 2 + d 4 ) n 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 d_5 = d_4(d_1+d_2+d_4) = 130 . But then 13 13 is a divisor of n n and d 4 < 13 < d 5 d_4 < 13 < d_5 . This case is impossible.


If d 4 = 11 d_4=11 then d 5 n = d 5 d 3 d 10 = d 5 d 3 ( d 1 + d 2 + d 4 ) d 8 = d 3 ( d 1 + d 2 + d 4 ) n 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 ) d_5 = d_3(d_1+d_2+d_4) .

  • If d 2 = 2 d_2=2 then d 1 + d 2 + d 4 = 14 d_1+d_2+d_4=14 divides n n , so that d 3 = 7 d_3=7 , and so d 5 = 98 d_5=98 . But then 49 49 divides n n and d 4 < 49 < d 5 d_4 < 49 < d_5 .
  • If d 2 = 3 d_2=3 then d 1 + d 2 + d 4 = 15 d_1+d_2+d_4=15 divides n n , so that d 3 = 5 d_3=5 , and so d 5 = 75 d_5=75 . But then 25 25 divides n n and d 4 < 25 < d 5 d_4 < 25 <d_5 .
  • If d 2 = 5 d_2=5 then d 3 = 7 d_3=7 and d 5 = 119 d_5=119 . But then 17 17 divides n n and d 4 < 17 < d 5 d_4 < 17 < d_5 .

None of these cases are possible


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


Thus we deduce that d 4 = 13 d_4=13 , and so n = ( d 1 + d 2 + d 4 ) d 8 n = (d_1+d_2+d_4)d_8 , so that d 5 = d 1 + d 2 + d 4 d_5 = d_1 + d_2 + d_4 ..

  • If d 2 = 2 d_2=2 then d 1 + d 2 + d 4 = 16 d_1+d_2+d_4=16 divides n n , so that 4 , 8 4,8 are also divisors of n n . This case is impossible.
  • If d 2 = 3 d_2=3 then d 1 + d 2 + d 4 = 17 d_1+d_2+d_4 =17 divides n n .
  • If d 2 = 5 d_2=5 then d 1 + d 2 + d 4 = 19 d_1+d_2+d_4=19 divides n n .
  • If d 2 = 7 d_2=7 then d 1 + d 2 + d 4 = 21 d_1+d_2+d_4=21 divides n n , so that 3 3 is a divisor of n n , which is impossible.
  • If d 2 = 11 d_2=11 then d 3 = 12 d_3=12 , which is impossible.

After running through all these cases, there are only six possible values of n n , namely 3 2 × 13 × 17 3^2\times13\times17 , 3 × 1 3 2 × 17 3\times13^2\times17 , 3 × 13 × 1 7 2 3\times13\times17^2 , 5 2 × 13 × 19 5^2\times13\times19 , 5 × 1 3 2 × 19 5\times13^2\times19 , 5 × 13 × 1 9 2 5\times13\times19^2 . Checking these cases shows that the correct solution is 3 2 × 13 × 17 = 1989 3^2 \times 13 \times 17 = \boxed{1989} .

There is a much faster way to see that d 4 is 13.
Since (d
1+d 2+d 4) (d_8)=d_( (d_4)-1) ) we know that (d_1+d_2+d_4) divides n which makes (d_1+d_2+d_4) a divisor.
So (d_1+d_2+d_4) = d_k for some k.
Since (d_1+d_2+d_4)>d_4 we know that k is at least 5.
Also since (d_1+d_2+d_4)

(d 8)=d ( (d 4)-1) ), we know that (d 1+d 2+d 4) (d_8) is a divisor of n and so (d_1+d_2+d_4) (d_8) <= n.

Though if k>=6 then (d 1+d 2+d 4)*(d 8) = (d k)*(d 8) > (d 5) * (d 8) = n which is a contradiction.
So k=5 and (d 1+d 2+d 4)*(d 8) = (d 5) * (d 8) = d 12 = d ( (d 4)-1) )
So d
4 - 1 = 12 which gives d_4 = 13.

Razzi Masroor - 1 year, 9 months ago

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Yes this is faster. By the way I'm suprised that the answer is the year!

Kelvin Hong - 1 year, 8 months ago

Your key observation is that d k = d 4 + d 2 + d 1 > d 4 d_k = d_4 + d_2 + d_1 > d_4 . Once we have that, we can slightly shorted your working to

n = d 12 d d 4 1 = ( d 1 + d 2 + d 4 ) d 8 d 5 d 8 = d 12 = n n = d_{12} \geq d_{d_4 - 1} = (d_1 + d_2 + d_4) d_8 \geq d_5 d_8 = d_{12} = n

Hence, we must have equality throughout, so d 4 1 = 12 d_4 - 1 = 12 .

Calvin Lin Staff - 1 year, 8 months ago

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