2025

Number

$2025$ (two thousand and twenty-five) is:

$3^4 \times 5^2$

$2025$ is $5$-smooth:

$\max \set {3, 5} = 5$

The sum of the first $9$ cubes:

$2025 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$

The $45$th square number after $1$, $4$, $9$, $16$, $25$, $36$, $\ldots$, $1296$, $1369$, $1444$, $1521$, $1600$, $1681$, $1764$, $1849$, $1936$:

$2025 = 45 \times 45$

Adding $1$ to each of its digits yields another square:

$2025 + 1111 = 3136 = 56^2$

The roots of those squares also differ by a repunit:

$45 + 11 = 56$

Not quite a Friedman number, but this is amusing:

$\sqrt {2025} = \paren {2 + 0!}^2 \times 5$

... and so is this:

$\sqrt {2025} = 20 + 25$

$\ds \map {\sigma_0} { 2025 }$

$=$

$\ds 15$

$\ds \map \phi { 2025 }$

$=$

$\ds 1080$

$\ds \map {\sigma_1} { 2025 }$

$=$

$\ds 7502$

1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $2025$
1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2025$