Find the positive integers n with exactly 1 2 divisors 1 = d 1 < d 2 < ⋯ < d 1 2 = n such that the divisor with index d 4 − 1 ( that is , d d 4 − 1 ) is ( d 1 + d 2 + d 4 ) d 8 .
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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.
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Yes this is faster. By the way I'm suprised that the answer is the year!
Your key observation is that d k = d 4 + d 2 + d 1 > d 4 . Once we have that, we can slightly shorted your working to
n = d 1 2 ≥ d d 4 − 1 = ( d 1 + d 2 + d 4 ) d 8 ≥ d 5 d 8 = d 1 2 = n
Hence, we must have equality throughout, so d 4 − 1 = 1 2 .
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Note that n = d j d 1 3 − j for any 1 ≤ j ≤ 1 2 . Since d d 4 − 1 = d 8 ( d 1 + d 2 + d 4 ) , we must have 8 < d 4 − 1 ≤ 1 2 , and hence 9 < d 4 ≤ 1 3 .
If d 4 = 1 0 then d 2 = 2 and 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 so that d 5 = d 4 ( d 1 + d 2 + d 4 ) = 1 3 0 . But then 1 3 is a divisor of n and d 4 < 1 3 < d 5 . This case is impossible.
If d 4 = 1 1 then d 5 n = d 5 d 3 d 1 0 = d 5 d 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 ) .
None of these cases are possible
If d 4 = 1 2 then 2 , 3 , 4 , 6 are all divisors of n , which is impossible.
Thus we deduce that d 4 = 1 3 , and so n = ( d 1 + d 2 + d 4 ) d 8 , so that d 5 = d 1 + d 2 + d 4 ..
After running through all these cases, there are only six possible values of n , namely 3 2 × 1 3 × 1 7 , 3 × 1 3 2 × 1 7 , 3 × 1 3 × 1 7 2 , 5 2 × 1 3 × 1 9 , 5 × 1 3 2 × 1 9 , 5 × 1 3 × 1 9 2 . Checking these cases shows that the correct solution is 3 2 × 1 3 × 1 7 = 1 9 8 9 .