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133114025660593 is a prime number
BaseRepresentation
bin11110010001000100000110…
…001101101101000010110001
3122110022120120022200021202111
4132101010012031231002301
5114421420011032114333
61151035423355045321
740016112435116035
oct3621040615550261
9573276508607674
10133114025660593
113946138323443a
1212b1a481a40841
1359377a2619473
1424c2a78b8c1c5
15105c9045284cd
hex79110636d0b1

133114025660593 has 2 divisors, whose sum is σ = 133114025660594. Its totient is φ = 133114025660592.

The previous prime is 133114025660579. The next prime is 133114025660701. The reversal of 133114025660593 is 395066520411331.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 81407816026384 + 51706209634209 = 9022628^2 + 7190703^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-133114025660593 is a prime.

It is a super-2 number, since 2×1331140256605932 (a number of 29 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (133114025660543) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66557012830296 + 66557012830297.

It is an arithmetic number, because the mean of its divisors is an integer number (66557012830297).

Almost surely, 2133114025660593 is an apocalyptic number.

It is an amenable number.

133114025660593 is a deficient number, since it is larger than the sum of its proper divisors (1).

133114025660593 is an equidigital number, since it uses as much as digits as its factorization.

133114025660593 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1749600, while the sum is 49.

The spelling of 133114025660593 in words is "one hundred thirty-three trillion, one hundred fourteen billion, twenty-five million, six hundred sixty thousand, five hundred ninety-three".