Esp if you knew the solution to the "Picking $n+1$ out of $2n$ numbers yields $a\mid b$", this solution is pretty natural. You just have to "do it" and check the details (A lot of multiplication involved) .
My writeup is very lengthy because it goes into the thinking and motivation. If you want to skip to the meat of the solution, read Brief Description and Case 6.
Let $L = (2l+1) \times 2^k$ be the integer less than 64 that is given. The variables $l, k$ will always refer to these values.
Note: I'm treating "less than 64" as "strictly less than 64", as that is advantageous to me. In particular, this solution doesn't work for the $L = 64$ case.
Recall the well-known variant of this problem: Recall the proof that for $n+1$ numbers chosen from $2n$, these must exists an $a \mid b$.
Define a set $S_m$ to have the divisible property if any 2 elements satisfy $a \mid b$.
Define a set $P_L$ to have the divisible-L property if any element $ a \neq L$ satisfies either $ a \mid L$ or $ L \mid a$.
Consider the $n$ sets of the form $S_{2s+1} = \{ (2s+1) 2^m | m \geq 0 \}$, where $s$ goes from $1$ to $n$. Each of these sets clearly has the divisible property (by a factor of $2^m$).
Then, given $n+1$ elements, by PHP there exists 2 that are in the same set, and hence divide each other.
We're in the scenario of $ 2n = 1992$.
We need $996+1 = 997$ elements, but are only given 994 elements. This means that we have to use the fact that $L < 64$ is an element to somehow reduce the number of sets by (at least) 3.
Brief description of the approach:
Step 1: Expand $S_{2l+1} $ (which consists of $L \times 2^m$ for some integer $m$) into $P_L$ by further adding in the multiples of $L$, so it has the divisible-L property. Note that I'm not including all possible factors of $L$.
- The terms that are left in the other $S_{2s-1} $ have the following form:
- No changes made. Group 1: $(2s-1 ) \times 2^b$ (for suitable values of $s$, namely $ (2l+1) \not \mid (2s-1)$. In particular, if $ l = 0$, then this set is empty.) It is unlikely that we can do better, so let's leave these sets aside.
- Removing multiple of $L$. Group 2: $(2l+1)(2s-1) \times 2^b$ for $s \neq 1, b \leq k-1 $. Now that some terms have been removed, how can we exploit this?
Step 2: Reorganize numbers in group 2 to reduce the number of sets by at least 3.
- For numbers of the form $s = (2l+1) \times x \times 3^a \times 2^b$, instead of having the sets $S_{(2l+1)\times x \times 3^a} $ with $b$ variable, we reorganize the sets to $ T_{(2l+1)\times x \times 2^b } $ with $a$ variable. Normally $b$ is large which is why we form the sets $S$, but having moved numbers with $b \geq k$ into $P_L$, it would be more optimal to group by $a$ when $a < k$ (which it likely for large enough $n$), thereby reducing the number of sets.
Solution: We consider cases based on the value of $k$.
Case 1: $L = (2l+1) \times 2^0$.
As it turns out, this is the easy case, but that makes the reasoning a slight red herring as the step 2 wasn't easily generalizable (See notes). We will still work through this as it's instructive.
Step 1: Previously $L$ was in $S_L$, and we should grow it into $P_L = \{ L, 2L, 3L, \ldots , \}$ which has the divisible-L property.
Step 2: Notice that $P_L = S_L \cup S_{3L } \cup S_{5L } \cup \ldots \cup S_{(2f-1)L } $, where $ 2f-1$ is the largest odd number $ \leq \frac{1992}{L}$. Since $ \frac{1992}{63} \geq 31$, hence $2f-1 \geq 31$. We've combined $f$ sets into 1, thus have reduced it by $f-1 \geq 15$ sets, and have at most $996 - 15 = 981$ sets.
By PHP, given 994 integers, at least 2 are in the same set, and we are done.
Musings: Since we only needed 982 elements, it suggest that when $L$ is a multiple of $ 2, 4, 8, 16, 32$, then it will get harder to reorganize the sets.
Let's now consider when $L = 1 \times 32$, which by this (shaky) reasoning should be the hardest case.
Case 6: $L = 1 \times 2^5$, the hardest case.
Step 1: Previously, $L$ was in $ S_1 = \{ 1, 2, \ldots, 32, \ldots 1024 \}$.
Let's grow it into $ P_{L} = \{ 1 \times 1, 1 \times 2, 1 \times 4 \ldots, 1 \times 32, 2 \times 32, 3\times 32, \ldots , 62 \times 32 \}$ to replace $S_1$. This has the divisible-L property.
The terms that are left in the other $S_{2s-1} $ have the form $(2s-1 ) \times 2^b$ for $ s \neq 1, 0 \leq b \leq 4$.
Step 2:
- Reorganize $ \{ s = 3 \times 3^a \times 2^ b | \, 0 \leq a \leq 5, 0 \leq b \leq 4, s \leq 1992 \}$ which are in $ S_3, S_9, S_{27}, S_{81}, S_{243}, S_{729} $, by placing them into $ T_{3 \times 2^b} = \{ t = 3 \times 3^a \times 2^b | \, 0 \leq a \leq 5, t \leq 1992 \}$, thereby reducing by 1 set.
- Likewise, reorganize $ \{ s = 5 \times 3^a \times 2^ b | \, 0 \leq a \leq 5, 0 \leq b \leq 4, s \leq 1992 \}$ which are in $ S_5, S_{15}, S_{45}, S_{135}, S_{405}, S_{1215} $, by placing them into $ T_{5 \times 2^b} = \{ t = 5 \times 3^a \times 2^b | \, 0 \leq a \leq 5, t \leq 1992 \}$, thereby reducing by 1 set.
- Likewise, reorganize $ \{ s = 7 \times 3^a \times 2^ b | \, 0 \leq a \leq 5, 0 \leq b \leq 4 , s \leq 1992 \}$ which are in $ S_7, S_{21}, S_{63}, S_{189}, S_{567}, S_{1701} $, by placing them into $ T_{7 \times 2^b} = \{ t = 7 \times 3^a \times 2^b | \, 0 \leq a \leq 5, t \leq 1992 \}$, thereby reducing by 1 set.
Hence, we've reduced by 3 sets, and so 994 terms is sufficient.
Case 2-5: $L = (2l+1) \times 2^k$ where $1 \leq k \leq 4$. We generalize the argument from Case 6, though the details are left to the reader to verify.
Step 1: Previously $L$ was in $ S_{2l+1}$.
We grow it into $ P_L \{ (2l + 1) \times 2, (2l +1) \times 4 , \ldots, L, 2L, 3L, \ldots \}$ to replace $S_{2l+1} $. This has the divisible-L property.
Step 2: Reorganize $ \{ s = (2l+1) \times x \times 3^a \times 2^ b |\, 0 \leq a \leq A, 0 \leq b \leq k-1 , s \leq 1992 \}$, by changing the indexing which reduces by $A-(k-1)$ sets. By setting $ x = 3, 5, 7$, in a similar calculation to Case 6, we can reduce by at least 3 sets.
Notes
- Case 1 Step 2 wasn't easily generalizable as the $k=0$ meant that there wasn't any Group 2 terms to look at. We had the full benefit of removing all of those sets resulting in the "combined $f$ sets into 1".
- Case 6 Step 2 also had $l=0$ which simplified the scenario a bit.
