7875

From ProofWiki
Jump to navigation Jump to search

Previous  ... Next

Number

$7875$ (seven thousand, eight hundred and seventy-five) is:

$3^2 \times 5^3 \times 7$


The total of all the entries in a magic cube of order $5$, after $1$, $36$, $378$, $2080$:
$7875 = \ds \sum_{k \mathop = 1}^{5^3} k = \dfrac {5^3 \paren {5^3 + 1} } 2$


The $1$st of the $8$th pair of triangular numbers whose sum and difference are also both triangular:
$7875 = T_{125}$, $8778 = T_{132}$, $8778 + 7875 = T_{182}$, $8778 - 7875 = T_{42}$


The $16$th odd abundant number after $945$, $1575$, $2205$, $2835$, $3465$, $4095$, $4725$, $5355$, $5775$, $5985$, $6435$, $6615$, $6825$, $7245$, $7425$:
$\map {\sigma_1} {7875} - 7875 = 8349 > 7875$


The $63$rd hexagonal number after $1$, $6$, $15$, $28$, $45$, $66$, $91$, $\ldots$, $5565$, $5778$, $5995$, $6216$, $6441$, $6670$, $6903$, $7140$, $7381$, $7626$:
$7875 = \ds \sum_{k \mathop = 1}^{63} \paren {4 k - 3} = 63 \paren {2 \times 63 - 1}$


The $125$th triangular number after $1$, $3$, $6$, $10$, $15$, $\ldots$, $6786$, $6903$, $7021$, $7140$, $7260$, $7381$, $7503$, $7626$, $7750$:
$7875 = \ds \sum_{k \mathop = 1}^{125} k = \dfrac {125 \times \paren {125 + 1} } 2$


Arithmetic Functions on $7875$

\(\ds \map {\sigma_1} { 7875 }\) \(=\) \(\ds 16 \, 224\) $\sigma_1$ of $7875$


Also see