Grant Sanderson
Region: US
Wednesday 25 March 2026 20:59:51 GMT
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Squill :
Pigeonhole principle says you can always find matching subsets
2026-03-27 02:29:30
0
Max Booth :
The optimum strategy for you is to pick 1,2,4,8,16,32,64 (then you’re out of options because anything else can be made from a combination of these), can’t justify yet though
2026-03-25 21:22:41
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pinkkonst :
he said subsets, not pairs, most comments got it wrong
2026-03-26 01:56:02
722
A❗❗ :
Pick 1,2,4,8,16,51,52,54,58, and 66. The goal is to use binary numbers, so the first 5 are powers of 2, and then you shift the numbers by 50 and do the same.
2026-04-03 02:12:58
3
tomerr :
there are 2^10=1024 subsets and sums are between 1 and 91+92+...+100 = 955 so by pigeonhole principle there are two subsets with the same sum, then remove elements that are in both
2026-03-26 13:36:07
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Foxy :
We are looking at a subset sum problem. After working on a piece of paper I was to find two sets that don't sum to themselves [1,2,4,8,16,32,64] and [100,99,98,96,92,84,68,36] with the addition of a single number to either set they brake. Therefore I have concluded that the winning strategy is to sum and not choose the numbers.
2026-03-26 21:01:08
7
Jellyfish Sandwich :
i'd say since there are 2¹⁰ subsets and the sum of those numbers can only be at best 91+92+93+94+95+96+97+98+99+100 < 1000 i have always a winning strategy
2026-03-25 21:14:38
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Ango 🤓🫱 :
Isn’t it fairly straightforward by the pigeonhole principle? 2^10 subsets, maximum possible sum in any of them is <1000<2^10, hence there’s some two sets with the same sum? (if they’re not disjoint just remove their intersection from both subsets and the remaining subset are now disjoint and still have equal sum)
2026-03-27 23:34:49
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لا اله الا الله :
1,2,4,7,11,16,22,29,37,46
2026-04-29 16:48:01
1
Exiv :
1, 3, 4, 5, 100, 97, 75, 67, 31, 22. Does that Work? Im not Sure
2026-04-21 00:35:00
0
Alex🇸🇰 :
2,3,5,7,11,13,17,19,23,29 wouldnt that work?
2026-04-25 23:33:09
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Le Yõ :
The maths to prove the strategy must be an absolute nightmare 😱… I can’t wait to dive into it!
2026-03-25 22:16:52
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Waerjak :
1 2 3 6 8 14 22 36 58 94
2026-03-28 23:06:10
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NoahMP3 :
isn't this just the first 10 odd numbers? 1,3,5,7,9,11,13,15,17,19
2026-04-21 04:31:04
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bobik :
well, you choose the numbers, so i guess you have the winning strategy. though it seems that there's a catch but i can't quite grasp it
2026-03-25 21:08:23
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Judith Radja Bologna :
In your examples you only presented disjoint subsets. I assume they are allowed not to be?
2026-03-26 17:00:28
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Max Booth :
The fact they are distinct is irrelevant since if we have an overlapped solution, removing the overlap from both solves it. Now we have 2^10-1 choices of sets, each ranging from 0-1000, and 2^10 - 1 > 1000 therefore pigeon hole principle, there are always two distinct sets with the same sum
2026-03-25 21:51:20
5
Waerjak :
1 2 3 6 9 15 24 39 63 102 (would be the first one with 10 numbers) but u need smaller than 100 so it's impossible
2026-03-31 08:12:57
1
tommyhspecial :
I'd rather play beer pong
2026-03-26 13:58:44
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Roie Busany :
Fibonacci (:
2026-03-25 22:32:36
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Lasse :
Ok when there would be two (not necessarily distinct) subsets with the same sum you could take out the intersection and have two distinct ones. So when there would be 10 numbers with no distinct subsets all subsets (2^10-1=1023) would have different sums. But all sums are in [1,1000) so contradiction
2026-04-16 22:52:49
1
⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻JOSHUA⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻⸻ :
fibonacci sequence, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89
2026-04-16 23:09:20
0
️ :
2 4 8 16 32 64 99 3 5 7
2026-03-28 02:51:36
0
sysomg :
Maybe I'm missing something, but the second person has no strategy. The first person has total control of who wins
2026-03-27 03:47:25
0
sandro :
Why not just choose 1-10?
2026-03-26 23:55:32
0
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