SUMaC 2025

Pre-College Topics Summer Programs
21

I am wondering the same.

22

Did anyone get an email about applying for Stanford Pre-Collegiate Summer Institutes? Does this mean anything or does anyone who applied to SUMAC get the email?

23

Emailed Stanford Pre-Collegiate Institutes to be on the safe side for this, here’s what they said:

Thank you for your email. Yes, you may now discuss the exam with other students or teachers however please do not publish the exam or exam contents on any social media platforms or distribute in any way.

So this means indirect discussions is okay as long as we don’t spoil the actual contents of the questions? Or outright ban on forums? Interpretation is a little iffy here.

24

when did you get the email? cuz I had a email around mid of feb saying I can apply to take a mathematics course through Stanford Pre-Collegiate Summer Institutes. I honestly don’t think it means anything

25

I got this yesterday. I was worried that maybe my solutions weren’t good enough which is why they suggested it. But I guess you’re right.

26

Well then I guess we can indeed discuss them. Ima just layout my strategies and how I tried to solve some of the problems:
Question 1 : Pretty simple, difference of squares and then Diophantine equations (basically equations with multiple integer solutions) and then pick the positive ones.
Question 2 : Part a) just show that a prime number appears down the sequence
Part b) Consider GCD(x,nx)=x. Then x/x + nx/x = n+1. So for all x =n+1, the set will be finite. For example 3 and 6, GCD(3, 2(3)) = 3 so 3/3+6/3 = 2+1 = 3, but 3 was already the seed so the set doesn’t grow.
Question 3 :- Part (A) This one was kinda tricky but basically you wanna notice that there is a third triangle in the middle that can be rotated to form the configuration from original to the one presented in part a. This allows us to prove that this configuration is indeed obtainable with the moves being XXYYXXYX.
Part(B) :- try that middle triangle rotation again and it doesn’t give you the configuration. Since NOTHING else was moved and this assumption worked for part A, by contra-positivity, part B shouldn’t be obtainable.
Question 4 :- Part (A) this one required me to use 10 pages of working
(I used strong induction)
Part(B) just show that the sum of the powers of 2s and added 1s will always pass over to the next number in the base (I basically used my idea of strong induction from part A in here)
Part 5: Part (A) - Brute force
Part (B) - No idea bruv
Part (C) - I showed an example of a (9,9) tour and basically argued that the first term and next term can’t share a common factor (also the reason why they can’t be even if the first term is even or 1 as well)
Question 6 - At this point I didn’t have much time left so I kinda just took a leap of faith and said for any n-points I said there are n-2 possible one lump snakes due to the restrictions.
Part (A) - 4 has 2, 5 has 3, 6 has 4 and 7 has 5
Part(B) - the pattern is cyclic so regardless of where you start, you will end up getting a snake if x-values are not equal to y-values and the values are not repeated on either x or y. So you only need to focus on the one lump condition but your apex point can’t be the points on the far ends. Thus removing 2 possibilities and hence showing that there only n-2 possible one lump snakes for n points.
These were my approaches on the problems and I have no idea if I got em right. Feel free to ask me, doubt me, correct me even. I might not respond fast but ill get to it eventually.
Good day

1 Like
27

A lot of our solutions align, so I’ll just note where we differ.

Question 3: a) I noticed the middle triangle pattern as well. I also made general comments about the puzzle as a directed graph and general behaviors. Got YYXYXX instead, however?
b) I just noted the simplest path that ONLY considered the minimum rotations to swap A and B, and figured that since the number of rotations 5 (mod 3) was not 0 [some mysterious shenanigans with triangle rotations/‘cycles’], it wasn’t possible.
Question 4: b) Proved via induction that alternating 1’s and 0’s was always unique, and each base was the sum of smaller bases, spent the most time on this and probably didn’t get it right.
Question 5: a) Brute force, however I did reduce the number of trees (of tours) to be drawn to 2 via some logic
b) Narrowed down number of trees to be drawn to 3, but didn’t have time to explore further. Vague comment about “there’s probably thousands of tours.”
c) Used (4,4) and (6,6) tours to say that there must be a k between 2 and n-2 and a (n-k) that are both co-prime to n for a (n,n) tour to exist (restriction on a_2 basically).
Question 6: a) Some brute forcing, albeit I did represent the problem as a graph to simplify the brute forcing. 4-2, 5-3, 6-5, 7-9. I shared your n-2 observation until I figured out the number of n=6 snakes.
b) Basically the order AFTER the peak n is predetermined, so it was a matter of figuring out how many combinations preceding the peak n there was. Did some manual adjustment to tighten the bound too. f(n) = ⌈ 1/n * (2^(n−1) − 1)⌉.

I feel solid about most of my solutions, and I hope I make it! Sort of poured all of my mathematical abilities into this.

28

Both YYXYXX and XXYYXXYX seem to work.
Could I see your of one-lump snakes for n=6. I wanna see what string did I miss. Honestly you seem really really enthusiastic about this, hope you get in. I had less than a month left when I started the admission test T_T. I think you have a much higher chance to get in. (You most likely will)(trust frfr)

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29

My solutions also seem to match both of yours. Maybe I didn’t do as bad as I thought.

2 Likes
30

For simpler notation, I’ll just note each (n, k_n) as a set containing just all k_n instead [ ex: 1,2),(2,3),(3,4),(4,1) becomes (2,3,4,1) ].

I got (4,6,5,3,2,1), (2,4,6,5,3,1), (2,3,5,6,4,1), (3,4,5,6,2,1), and (2,3,4,5,6,1). What you might’ve missed is the ‘double’ peaks at n=4 since two of the snakes have their peaks there.

Thank you! I basically camped the website daily in December and started working the moment applications opened (especially since this is the only math camp I have time for over the summer
).

1 Like
31

I started 20 days before the application was due :grimacing:. Hope to meet you if I get admitted. Thank you for showing me my mistakes.

1 Like
32

i just received an email saying that i should apply fort he pre-college institute and they said they would give me a waiver to the application fee, does that mean I dont have a chance w/ sumac??

33

I got this email as well. Did you list that you applied for other summer programs? I’m hoping they sent that to me because I didn’t add any.

34

What time did you receive it and did you apply to any other Stanford summer programs?

If you received this email early in the morning then that could be an automated email to all applicants or few who didn’t apply to other available summer programs.

That said, this email was sent to rejected or waitlisted students in the past but only after decisions were released. I don’t know if this is an early indicator though. Not everyone checks their spam folders often so who knows?

Check this old thread for reference.

They use the same admission/application system and email for communication so that makes things easier on their end to target certain students if they chose to.

35

I got it at 3:00 AM hoping it is automated.

36

3:00 AM PDT? then it could be but who knows.

It doesn’t matter, keep your options open and keep doing good work. this program doesn’t make or break you.

37

gg problem 6b bound is based xd

i found a weird and bad recursion approach that gave 3^{n-3} but then i realized the obvious approach had 2^{n-2} so i wrote it up anyway and my thoughts :skull:

how u got the bound?

Also on problem 4 I did inductive approach too and got that 2^k+(k+1) is unique expressible if k is uniquely expressible also
 idk if that confirm or deny ur claim ;-;

i didnt notice the middle triangle thing like u and eureka and found a goofy casework reduction sol ;-;

im cooked

38

same, i think it is automated

i need to see my sumac decision aaaaa

39

The 2^(n-1)-1 part came from me completely overlooking the obvious restriction of 2^(n-2). I did a summation from i=2 to n-1 of (n-1) choose (i-1) to basically simulate the remaining combinations of orders of the remaining numbers AFTER peak value n (and since 1 is obviously only the last value, it’s not in the range of i). I think I slipped up in my arithmetic somewhere, since I missed the 2^(n-2) and instead got 2^(n-1)-1.

From that point on, I basically just played around with various ideas, and I randomly stumbled upon f(n)/actual-value being about equal to n, so I just added the 1/n to minimize as much error as possible. Then I realized that for some values of n, f(n) undershot slightly so I did the ceiling function to address that. It was basically a game of wild goose chase after the summation.

I think your problem 4 approach doesn’t interfere with any of mine (in fact, it seems to overlap with my own inductive proof of alternating 1’s and 0’s!), so you should be good there.

Honestly wishing decisions weren’t a whole month away, but I’ll concern myself with the school grind in the meantime :flexed_biceps:

40

yo anyone get 5b?
i think that may be close to impossible without excessive bruteforce ngl
but ill be impressed if someone has a solution