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135313142543099 is a prime number
BaseRepresentation
bin11110110001000100001011…
…110010110111001011111011
3122202002210212120001210202122
4132301010023302313023323
5120213432303442334344
61155441555521551455
740334023546155413
oct3661041362671373
9582083776053678
10135313142543099
113a129a883a5154
12132147157a558b
135a66c89237771
14255b296319643
151099c13185aee
hex7b110bcb72fb

135313142543099 has 2 divisors, whose sum is σ = 135313142543100. Its totient is φ = 135313142543098.

The previous prime is 135313142543003. The next prime is 135313142543101. The reversal of 135313142543099 is 990345241313531.

It is a strong prime.

It is an emirp because it is prime and its reverse (990345241313531) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-135313142543099 is a prime.

Together with 135313142543101, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 135313142543099.

It is not a weakly prime, because it can be changed into another prime (135313142513099) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67656571271549 + 67656571271550.

It is an arithmetic number, because the mean of its divisors is an integer number (67656571271550).

Almost surely, 2135313142543099 is an apocalyptic number.

135313142543099 is a deficient number, since it is larger than the sum of its proper divisors (1).

135313142543099 is an equidigital number, since it uses as much as digits as its factorization.

135313142543099 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 5248800, while the sum is 53.

The spelling of 135313142543099 in words is "one hundred thirty-five trillion, three hundred thirteen billion, one hundred forty-two million, five hundred forty-three thousand, ninety-nine".