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mretka :heart_cyber:​

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

20 comments
Rich Holmes

@mretka Won't see its like for a thousand years!

Juan Escamilla Mólgora :gauss:

@mretka

Let’s also add that null element!

\[ 2025 = 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 \]

Smoljaguar

@mretka Also, 2025=45^2=(1+2+3+4+5+6+7+8+9)^2 (this is not a coincidence!!)

éric 🚲 🇪🇺 :emacs:

@mretka And only happens two more times before the year 10,000, in the years 3025 and 9801!

Natasha Nox 🇺🇦🇵🇸

@ericsfraga @mretka Although we might use the calendar of human history by then instead of Gregorian and skip those years. 🫠

Rich Holmes

@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

Σ(i³) = (Σi)²

@mretka Also works for 2024 if you include -1³+0³+1³... 🤷‍♂️

Mathias Micheel

@mretka oh, 2025 I'd also 45 squared, and I wanted to give my brother who turns 45 a giant square as a present made from smaller squares... Maybe I will add the cubes as well😅

David C. Norris

@mretka Trust but verify ...
julia> sum((1:9).^3)
2025

Walter Tross

@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")

I am Jack's Lost 404

@mretka

Also, starting in 2025, pi will be exactly equal to 3 :blobupsidedown:

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