2025

From ProofWiki
Jump to navigation Jump to search

Previous  ... Next

Number

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

$3^4 \times 5^2$


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:
$45 + 11 = 56$


Also see


Sources