2025: Division by digit removal.

[Alan Mackenzie]

The year is now 2025.

2025 is divisible by 9. To do this, you merely have to remove the digit
0, leaving 225.

225 is also divisible by 9. Again, remove one of the 2 digits, leaving
25.

25 is divisible by 5. Guess what, to do this you remove the 2 digit
leaving 5.

Well, that's as far as it goes.

Happy new year to you all!

[Jim Burns]

Post by Alan Mackenzie
The year is now 2025.
2025 is divisible by 9.
To do this,
you merely have to remove the digit 0,
leaving 225.
225 is also divisible by 9.
Again, remove one of the 2 digits,
leaving 25.
25 is divisible by 5. Guess what,
to do this you remove the 2 digit
leaving 5.
Well, that's as far as it goes.

Once more:
5 is divisible by nothing (is prime).
Remove the 5,
leaving nothing.

Post by Alan Mackenzie
Happy new year to you all!

Happy New Year!

[sobriquet]

Post by Alan Mackenzie
The year is now 2025.
2025 is divisible by 9. To do this, you merely have to remove the digit
0, leaving 225.
225 is also divisible by 9. Again, remove one of the 2 digits, leaving
25.
25 is divisible by 5. Guess what, to do this you remove the 2 digit
leaving 5.
Well, that's as far as it goes.
Happy new year to you all!

https://www.wolframalpha.com/input?i=%28sum+n%3D1+to+9+n%29%5E2

https://www.wolframalpha.com/input?i=+sum+n%3D1+to+9+n%5E3

[Moebius]

Post by sobriquet

Post by Alan Mackenzie
The year is now 2025.
2025 is divisible by 9.  To do this, you merely have to remove the digit
0, leaving 225.
225 is also divisible by 9.  Again, remove one of the 2 digits, leaving
25.
25 is divisible by 5.  Guess what, to do this you remove the 2 digit
leaving 5.
Well, that's as far as it goes.
Happy new year to you all!

https://www.wolframalpha.com/input?i=%28sum+n%3D1+to+9+n%29%5E2
https://www.wolframalpha.com/input?i=+sum+n%3D1+to+9+n%5E3

Jeah,

2025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 1^3 + 2^3 + 3^3 + 4^3 +
5^3 + 6^3 + 7^3 + 8^3 + 9^3.

[Moebius]

Post by Moebius

Post by sobriquet

Post by Alan Mackenzie
The year is now 2025.
2025 is divisible by 9.  To do this, you merely have to remove the digit
0, leaving 225.
225 is also divisible by 9.  Again, remove one of the 2 digits, leaving
25.
25 is divisible by 5.  Guess what, to do this you remove the 2 digit
leaving 5.
Well, that's as far as it goes.
Happy new year to you all!

https://www.wolframalpha.com/input?i=%28sum+n%3D1+to+9+n%29%5E2
https://www.wolframalpha.com/input?i=+sum+n%3D1+to+9+n%5E3

Jeah,
2025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 1^3 + 2^3 + 3^3 + 4^3 +
5^3 + 6^3 + 7^3 + 8^3 + 9^3.

1729.

"1729 is also known as Ramanujan number or Hardy–Ramanujan number, named
after an anecdote of the British mathematician G. H. Hardy when he
visited Indian mathematician Srinivasa Ramanujan who was ill in a
hospital.[14][15] In their conversation, Hardy stated that the number
1729 from a taxicab he rode was a "dull" number and "hopefully it is not
unfavourable omen", but Ramanujan remarked that 'it is a very
interesting number; it is the smallest number expressible as the sum of
two cubes in two different ways'." (Wikipedia)

See: https://en.wikipedia.org/wiki/Taxicab_number

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