793

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Number

$793$ (seven hundred and ninety-three) is:

$13 \times 61$

The $6$th and last in a sequence of $6$ consecutive integers which are divisible by the corresponding term in the sequence of the first $6$ primes:

$793 = 13 \times 61$

Also see

• Previous: Sequence of Numbers Divisible by Sequence of Primes

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $788$