2018-12-31
[public] 112K views, 5.75K likes, 66.0 dislikes audio only
Huge thanks to Inder Taneja for crunching the numbers. I asked for a few number facts: they produced a whole paper.
Inder's work on their site: https://inderjtaneja.com/2018/12/31/2019-in-numbers/
The full PDF: https://zenodo.org/record/2529103
Here are text version you can copy of all the facts in the video:
2019 =
(2×22+2/2)^2 −2−2−2
1+(1+1)^11 − (11−1)×(1+1+1)
9+(9999−9) / 9+9×99+9
(aaaaa−a)×(a+a)−a×aa / a×aa
1+2×34+5×6×(7×8+9)
98+7+65+43^2×1
−1×(2+3)+√[4^(5+6)]−7−8−9
√9+8−7+65×(4+3^(2+1))
(1+2)×(3!!−4!−5+6−7−8−9)
9+(8+7)×(6!/5−4−3−2−1)
0^3 + 1^8 + 2^7 − 3^9 + 4^6 + 5^4 + 6^2 + 7^5 + 8^1 + 9^0
1^4 + 2^4 + 3^4 + 5^4 + 6^4
√[1155^2 + 1656^2]
CORRECTIONS
- I've just noticed the descending square-root sequence still uses concatenation. Sorry. If you find a better version give me a shout.
- Torin Storkey was the first to suggest the solution for zero is: 0! + 0! + 0! + 0! + 0! + 0! + … + 0! = 2019
- Let me know if you spot anything else!
For the first time: I'm using the frame from the video youtube automatically picked as the default thumbnail. I can't do better than that.
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