2601

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Number

$2601$ (two thousand, six hundred and one) is:

$3^2 \times 17^2$

Can be expressed as the sum of $3$ squares in $12$ different ways.

\(\ds \quad \ \ \)

\(\ds 2601\)

\(=\)

\(\ds 1^2 + 10^2 + 50^2\)

\(\ds \)

\(=\)

\(\ds 2^2 + 14^2 + 49^2\)

\(\ds \)

\(=\)

\(\ds 10^2 + 10^2 + 49^2\)

\(\ds \)

\(=\)

\(\ds 14^2 + 14^2 + 47^2\)

\(\ds \)

\(=\)

\(\ds 1^2 + 22^2 + 46^2\)

\(\ds \)

\(=\)

\(\ds 14^2 + 17^2 + 46^2\)

\(\ds \)

\(=\)

\(\ds 1^2 + 34^2 + 38^2\)

\(\ds \)

\(=\)

\(\ds 14^2 + 31^2 + 38^2\)

\(\ds \)

\(=\)

\(\ds 3^2 + 36^2 + 36^2\)

\(\ds \)

\(=\)

\(\ds 24^2 + 27^2 + 36^2\)

\(\ds \)

\(=\)

\(\ds 17^2 + 34^2 + 34^2\)

\(\ds \)

\(=\)

\(\ds 22^2 + 31^2 + 34^2\)

The $17$th pentagonal pyramidal number after $1$, $6$, $12$, $40$, $75$, $126$, $196$, $288$, $405$, $550$, $726$, $936$, $1183$, $1470$, $1800$, $2176$:

$2601 = \ds \sum_{k \mathop = 1}^{17} \dfrac {k \paren {3 k - 1} } 2 = \dfrac {17^2 \paren {17 + 1} } 2$

The $51$st square number after $1$, $4$, $9$, $16$, $25$, $36$, $\ldots$, $1521$, $1600$, $1681$, $1764$, $1849$, $1936$, $2025$, $2116$, $2209$, $2304$, $2401$, $2500$:

$2601 = 51 \times 51$

Also see

• 2601 as Sum of 3 Squares in 12 Different Ways

• Previous  ... Next: Pentagonal Pyramidal Number
• Previous  ... Next: Square Number

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2601$