What is the max of $n$ such that
$$\sum_{i=1}^n\frac{1}{a_i}=1$$
where $a_{i}\ (i=1,2,\cdots,n)$ are integers which satisfy $2\le a_1\lt a_2\lt\cdots\lt a_n\le 99$ ?
Also, I need how to prove that the $n$ you get is the maximum.
My approach:
I'm going to represent $\sum_{i=1}^n\frac{1}{a_i}$ as $(a_1,a_2,\cdots,a_n).$
$(2,3,6)\rightarrow(4,5,7,9, 12, 15,18,30,42,45,90)$
$\rightarrow(8,9,10, 12, 14, 15, 16, 18,22,27,
30, 35, 40, 42, 45, 48, 54, 56, 60, 72,
90, 99)$
(here I used $(3)=(4,12), (6)=(7,42)$ etc.)
This is the $n=22$ case.
Update: I've just got the following $n=42$ case. I don't know if this is the max.
$(8,9,10, 12, 14, 15, 16, 18,22,27,
30, 35, 40, 42, 45, 48, 54, 56, 60, 72,
90, 99)\rightarrow(15,17,20,21,22,26,27,30,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60,63,66,70,75,76,77,78,80,84,85,88,90,91,95,96,99)$
Here I used $(10)=(17,34,85), (14)=(28,44,77)$ etc.
Update 2 : I've just got another $n=42$ case.
$(17,18,20,21,22,24,26,27,32,33,34,35,36,38,39,40,42,44,45,48,50,52,54,55,56,60,63,66,70,72,75,76,77,78,80,84,85,88,91,95,96,99)$
Here I used $(15,30,90)=(18,24,72)$.
Update 3 : I crossposted to MO.
Best Answer
[EDITED to include other prime powers and give the list of 27 maximal solutions]
The maximum is $42$, attained in $27$ ways (listed below); there are $566$ runners-up with $41$, and then $6747$ with $40$, another $52078$ with $39$, "etc.".
The counts are obtained by dynamical programming. We simplify the computation by checking that the $a_i$ can include no multiple of a prime power greater than $27$, and if multiples of $11$, $13$, $16$, $17$, $19$, $25$, or $27$ appear then they must combine to remove that factor from the denominator, which can only be done in $46,\phantom. 9,\phantom. 7,\phantom., 1, \phantom. 2,\phantom. 1,\phantom. 1$ ways respectively (the last four are $(17,34,85)$, $(19,57,76)$ or $(38,76,95)$, $(50,75)$, and $(25,54)$ respectively; $23$ does not occur). That brings the denominator down to $D = 2^3 3^2 5 \phantom. 7 = 2520$, small enough to make a table of the number of times each pair of integers arises as $(n, D\sum_{i=1}^n 1/a_i)$ with $\sum_i 1/a_i \leq 1$, and at the end extract the counts for $(n,D)$.
This approach does not immediately give the list of $27$ maximal solutions, but it can be modified to compute this list instead: at each stage, instead of recording the number of representations of each fraction, keep track of the representation(s) with the largest number of terms. The list of $42$'s is follows, in lexicographical order; each uses the prime 17, and all use 99 except for two which have $\max_i a_i = 96$.