125,000

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Number

$125 \, 000$ (one hundred and twenty-five thousand) is:

$2^3 \times 5^6$

The $50$th cube number

$125 \, 000 = 50 \times 50 \times 50$

Also see

• Previous  ... Next: Cube Number

Historical Note

The second case of Fermat's Last Theorem, where the value $p$ in the equation $x^p + y^p = z^p$ divides one of $x$, $y$ and $z$, had been proved impossible for all values of $p$ up to $125\,000$, at the point at which Andrew Wiles finally proved it in $1994$.

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $125,000$