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135055536433 = 478192824307
BaseRepresentation
bin111110111000111101…
…1111111000100110001
3110220121020211002211221
41331301323333010301
54203043204131213
6142013232505041
712520542055426
oct1756173770461
9426536732757
10135055536433
1152305369514
12222119ab181
13c9743a97cc
146772aa5d4d
1537a6ada38d
hex1f71eff131

135055536433 has 4 divisors (see below), whose sum is σ = 135058408560. Its totient is φ = 135052664308.

The previous prime is 135055536431. The next prime is 135055536439. The reversal of 135055536433 is 334635550531.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 135055536433 - 21 = 135055536431 is a prime.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 135055536433.

It is not an unprimeable number, because it can be changed into a prime (135055536431) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1364335 + ... + 1459972.

It is an arithmetic number, because the mean of its divisors is an integer number (33764602140).

Almost surely, 2135055536433 is an apocalyptic number.

It is an amenable number.

135055536433 is a deficient number, since it is larger than the sum of its proper divisors (2872127).

135055536433 is an equidigital number, since it uses as much as digits as its factorization.

135055536433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2872126.

The product of its (nonzero) digits is 1215000, while the sum is 43.

The spelling of 135055536433 in words is "one hundred thirty-five billion, fifty-five million, five hundred thirty-six thousand, four hundred thirty-three".

Divisors: 1 47819 2824307 135055536433