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Math
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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
This is pretty silly. This is the 5th fermat number, which is 232+1. It can be written as 228(54+24)−(5⋅27)4+1=228⋅641−(6404−1), so it's clear that 641 divides it. For another prime that divides it, we must have 232≡−1(modp), and therefore 264≡1(modp), so by Fermat's Little Theorem we must have that 64 divides p-1, or that p=64k+1. Now there are 9 possibilities, but some of them aren't prime. It's easy to check them.
6 points for Patrick Hompe! -7 points for calling it silly! >:O Did you know that in general, each prime factor of Fn≥2 is in the form of k⋅2n+2+1,k∈N?
def primes(n):
if n<=2:
return []
sieve=[True]*(n+1)
for x in range(3,int(n**0.5)+1,2):
for y in range(3,(n//x)+1,2):
sieve[(x*y)]=False
return [2]+[i for i in range(3,n,2) if sieve[i]]
Why does this have the "points exchange" tag? Also, is this possible to do without W|A? If somebody finds it without W|A, it's pure luck. Also, fun fact: my FTC Robotics team ID number is 7297.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
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or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
This is pretty silly. This is the 5th fermat number, which is 232+1. It can be written as 228(54+24)−(5⋅27)4+1=228⋅641−(6404−1), so it's clear that 641 divides it. For another prime that divides it, we must have 232≡−1(modp), and therefore 264≡1(modp), so by Fermat's Little Theorem we must have that 64 divides p-1, or that p=64k+1. Now there are 9 possibilities, but some of them aren't prime. It's easy to check them.
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+1 for the first sentence.
6 points for Patrick Hompe! -7 points for calling it silly! >:O Did you know that in general, each prime factor of Fn≥2 is in the form of k⋅2n+2+1, k∈N?
641
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Python:
Euler showed that this is the first composite Fermat number :D 641 divides it
Also, all the prime factors of a2n+b2n are of the sort 1+k.2n+1
Why does this have the "points exchange" tag? Also, is this possible to do without W|A? If somebody finds it without W|A, it's pure luck. Also, fun fact: my FTC Robotics team ID number is 7297.
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Yes, there's a brilliant solution to this. The answer is 641, so you get the 1 point.
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the 1 point or 7 points?
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"Brilliant solution"
Lol pun