The New Sudoku Players' Forum

[Serg]

Hi, people!
My friend sent me amazing mathematical observations (I don't know who is the author of these formulae). Sign "^" denotes "power" sign.

2025 = (20 + 25)^2
2025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2
2025 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3

Serg

[rjamil]

Perfect square:
The year 2025 holds a unique mathematical distinction as a "perfect square year," being the square of 45 (45 x 45 = 2025). This is a relatively rare occurrence, with the last one being 1936 (44 x 44) and the next one being 2116 (46 x 46).

Sum of Three Squares:
40² + 20² + 5² = 2025

Product of Squares:
9² * 5² = 2025

Harshad Number:
(also known as a Niven number), meaning it is divisible by the sum of its digits (2+0+2+5 = 9),

R. Jamil

[Serg]

Hi, Jamil!
Property "2025 = (20 + 25)^2" is extremely rare. Let's call such years, having even number of digits, "balanced". It's turn out, there are only 3 such 4-digit years: 2025, 3025, 9801. Not every millenium contains such "balanced" years.

Serg

[rjamil]

2025 as the 23rd Centered Octagonal Number:
These numbers represent a geometric pattern where a point is placed at the center, and subsequent layers of dots form an octagon around it.

The formula for the nth centered octagonal number is 1 + 8 * (n * (n - 1) / 2). For n=23, this formula calculates to 1 + 8 * (22 * 23 / 2) = 1 + 8 * 253 = 2025.

R. Jamil

[Leren]

I think you have missed an "important" one. It's part of a Pythagorean Triple (sum of two squares and is a square) : 27^2 + 36^2 = 45^2 = 2025.

or even maybe the sum of a power of six and one of 4 and is a square : 3^6 + 6^4 = 45^2 = 2025

or even 2025 = 401 + 403 + 405 +407 +409.

In date form two dates in the year 2025 i.e. 9/12/2025 or 12/9/2025 form part of a Pythagorean Quadruple : 9^2 + 12^2 + 20^2 = 25^2

Other Pythagorean dates in 2025 are 24/7/2025, just a few days ago, which is part of the Pythagorean Triple 7^2 + 24^2 = 25^2 - a so called limping square Pythagorean Triple.

or in about a month's time 4/9/25 is related to the most well known Pythagorean Triple of them all 3^2 + 4^2 = 5^2.

Leren

[rjamil]

The Prime Factorization:
3 x 3 x 3 x 3 x 5 x 5 = 3⁴ × 5² = 2025

R. Jamil

[Cenoman]

2025 = (20 + 25)^2
2025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2
2025 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3
Serg

The first formula sets 2025 as a Kaprekar number (Kaprekar: Indian mathemathician, 20th century)
The following two formulae are the result of Nicomachus theorem: for any integer n, square of the sum (1 + ... + n) = sum of the cubes
Previous such year was 1296, next will be 3025. (Nicomachus of Gerasa: Pythagorean mathemathician, 2nd century. Gerasa, now Jerash, in Jordan)

Also, not mentionned in above posts: 2025 = 1 + 3 + ... + 89 (sum of the 45 first odd numbers)
Will be true also for the next perfect square year, 2116 (sum of the 46 first odd numbers)

[Serg]

Hi, Cenoman!

2025 = (20 + 25)^2
2025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2
2025 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3
Serg

The first formula sets 2025 as a Kaprekar number (Kaprekar: Indian mathemathician, 20th century)
The following two formulae are the result of Nicomachus theorem: for any integer n, square of the sum (1 + ... + n) = sum of the cubes
Previous such year was 1296, next will be 3025. (Nicomachus of Gerasa: Pythagorean mathemathician, 2nd century. Gerasa, now Jerash, in Jordan)

Also, not mentionned in above posts: 2025 = 1 + 3 + ... + 89 (sum of the 45 first odd numbers)
Will be true also for the next perfect square year, 2116 (sum of the 46 first odd numbers)

Thanks for interesting references! Amaizing, that
3025 = (30 + 25)^2
3025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)^2
3025 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 + 10^3

Serg