For each integer $n = p_1p_2p_3p_4$, where $p_1,p_2,p_3,p_4$ are distinct primes.
Let $(d_1=1) \lt d_2 \lt d_3 \lt \cdots \lt d_{15}\lt (d_{16} = n)$ be $16$ positive divisors of $n$.
Prove that if $n \lt 1995$ then $d_9-d_8 \not = 22$.

My Work
Without loss of generality assume $p_1 \lt p_2 \lt p_3 \lt p_4$. Then it is easy to see that $d_9-d_8 = p_2p_3 - p_1p_4$
Now how to continue from here?

Source: This is a problem from 1995 Irish Mathematical Olympiad.

  • $\begingroup$ Are the $d_i$ the 16 divisors of $n$? The name and your statement suggests that, but it's not written anywhere. $\endgroup$ – martini Feb 1 '17 at 8:42
  • $\begingroup$ @martini opps... I missed that.. editing $\endgroup$ – Rezwan Arefin Feb 1 '17 at 8:43
  • 1
    $\begingroup$ Actually there are not that many choices for $p_1$ and $p_4$ when $n<1995$. We have $p_1 \in \{2,3\}$ and $p_4 \in \{7,11,13,17\}$. You can just brute force it from here...Note that $p_1=2$ leads to an odd result, hence you just have to check $p_1=3$.. $\endgroup$ – MooS Feb 1 '17 at 8:51

If $d_9-d_8=p_2p_3-p_4p_1$, then suppose it's equal to 22 then clearly p cannot be 2 and since $n<1995$ it follows that $p_4<19$.
Now if you suppose that 3 is not on of primes then you get $n>1995$, same happens for 5 and 7 so $p_1=3,p_2=5,p_3=7$ and now you are left only with $p_4=11,13,17$, so check cases.


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