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493399546973 is a prime number
BaseRepresentation
bin1110010111000001110…
…01111001110001011101
31202011112212021221212022
413023200321321301131
531040440241000343
61014355320345525
750434620644144
oct7134071716135
91664485257768
10493399546973
111802821aa443
127b75a6152a5
13376b1858178
1419c4879075b
15cc7b43d568
hex72e0e79c5d

493399546973 has 2 divisors, whose sum is σ = 493399546974. Its totient is φ = 493399546972.

The previous prime is 493399546909. The next prime is 493399546979. The reversal of 493399546973 is 379645993394.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 365998380484 + 127401166489 = 604978^2 + 356933^2 .

It is a cyclic number.

It is not a de Polignac number, because 493399546973 - 26 = 493399546909 is a prime.

It is a super-2 number, since 2×4933995469732 (a number of 24 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 493399546897 and 493399546906.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (493399546979) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 246699773486 + 246699773487.

It is an arithmetic number, because the mean of its divisors is an integer number (246699773487).

Almost surely, 2493399546973 is an apocalyptic number.

It is an amenable number.

493399546973 is a deficient number, since it is larger than the sum of its proper divisors (1).

493399546973 is an equidigital number, since it uses as much as digits as its factorization.

493399546973 is an evil number, because the sum of its binary digits is even.

The product of its digits is 595213920, while the sum is 71.

The spelling of 493399546973 in words is "four hundred ninety-three billion, three hundred ninety-nine million, five hundred forty-six thousand, nine hundred seventy-three".