Just learned this #mathstodon #math
2024 = 2³+3³+4³+5³+6³+7³+8³+9³
And for next year we'll have the entire sum of cubes from 1 to 9!
2025 = 1³+2³+3³+4³+5³+6³+7³+8³+9³
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@mretka Won't see its like for a thousand years!
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@mretka Also, 2025=45^2=(1+2+3+4+5+6+7+8+9)^2 (this is not a coincidence!!)
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@mretka
And, for this next year, (20+25)²=2025.
[#]mathstodon #math
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@ericsfraga Wow! :mind_blown: :mind_blown:
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@mretka And only happens two more times before the year 10,000, in the years 3025 and 9801!
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@ericsfraga @mretka Although we might use the calendar of human history by then instead of Gregorian and skip those years. 🫠
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@Natanox @mretka True; this is all rather calendar dating (culturally) dependent!
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@ericsfraga @mretka Interesting that:
2025 = 1³+2³+3³+4³+5³+6³+7³+8³+9³
3025 = 1³+2³+3³+4³+5³+6³+7³+8³+9³+10³
and
2025 = (20+25)²
3025 = (30+25)²
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@doctroid @mretka Good point. Well spotted! Love it.
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@ericsfraga @mretka
So the world's ending next year then right?
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@mretka
Remarkable and
a remarkable discovery!
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@mretka @pozorvlak Wow, two years in a row! What are ths odds?
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@mretka And, by a known equality, also
2025 = (1+2+3+4+5+6+7+8+9)²
(The equality can be stated as "the sum of the cubes of the integers in a range starting from 0 or 1 is the square of the sum of the same integers")
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@mretka
Also, starting in 2025, pi will be exactly equal to 3 :blobupsidedown:
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