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268 sats \ 6 replies \ @south_korea_ln 23 Oct 2024 \ on: Math Puzzle: SimpleStacker Edition science
Focus on the right-hand side, we can distribute the first factor (2^a - 1) into the second one:
= 2^a \cdot \left( 2^{a(b-1)} + 2^{a(b-2)} + \ldots + 2^a + 1 \right) - \left( 2^{a(b-1)} + 2^{a(b-2)} + \ldots + 2^a + 1 \right).
The first term gives
2^a \cdot \left( 2^{a(b-1)} + 2^{a(b-2)} + \ldots + 2^a + 1 \right) = 2^{ab} + 2^{a(b-1)} + \ldots + 2^{2a}
and the second term gives
- \left( 2^{a(b-1)} + 2^{a(b-2)} + \ldots + 2^a + 1 \right)
= - 2^{a(b-1)} - 2^{a(b-2)} - \ldots - 2^a - 1
When adding up these two terms, it's easy to see that all terms but the first and the last cancel out proving the identity.
So this basically proves that one Mersenne number (which isn't prime) can be factored out into another Mersenne number. And it also proves that if the exponent is of the form ab it cannot be a Mersenne prime.
Happy to see other people post puzzles as well :)
Anyone with time and coding skills, check out this issue to add spoiler formatting in SN. In addition to the usual SN contribution reward, I am giving out a 30k sat bounty.
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21 sats \ 0 replies \ @south_korea_ln 23 Oct 2024
#736461
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0 sats \ 3 replies \ @k00b 23 Oct 2024
Curious: would it be better to left align math or do you prefer it centered?
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343 sats \ 1 reply \ @SimpleStacker OP 23 Oct 2024
It can be controlled by the user
y=f(x)
y=f(x)
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0 sats \ 0 replies \ @south_korea_ln 23 Oct 2024
Oh i didn't realize that. But makes sense after i checked how you did it.
$$y = f(x)$$
vs
$$
y = f(x)
$$
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294 sats \ 0 replies \ @south_korea_ln 23 Oct 2024
Centered is fine. That's what I'm used to in my publications anyhow. Don't have a strong opinion on this.
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0 sats \ 0 replies \ @SimpleStacker OP 23 Oct 2024
Good job :)
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