Base | Representation |
---|---|
bin | 101011100010101111000… |
… | …0001111001000010111001 |
3 | 210012002221210110021010212 |
4 | 1113011132001321002321 |
5 | 1241022142022404213 |
6 | 20421120504520505 |
7 | 1155235446112331 |
oct | 127053601710271 |
9 | 23162853407125 |
10 | 5984466997433 |
11 | 19a7aa8672751 |
12 | 8079b5a80135 |
13 | 345443275112 |
14 | 169914ca20c1 |
15 | a5a0a48b9a8 |
hex | 5715e0790b9 |
5984466997433 has 4 divisors (see below), whose sum is σ = 5984501956608. Its totient is φ = 5984432038260.
The previous prime is 5984466997403. The next prime is 5984466997511. The reversal of 5984466997433 is 3347996644895.
5984466997433 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
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 a de Polignac number, because none of the positive numbers 2k-5984466997433 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (5984466997403) 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, 17221541 + ... + 17565602.
It is an arithmetic number, because the mean of its divisors is an integer number (1496125489152).
Almost surely, 25984466997433 is an apocalyptic number.
It is an amenable number.
5984466997433 is a deficient number, since it is larger than the sum of its proper divisors (34959175).
5984466997433 is a wasteful number, since it uses less digits than its factorization.
5984466997433 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 34959174.
The product of its digits is 4232632320, while the sum is 77.
The spelling of 5984466997433 in words is "five trillion, nine hundred eighty-four billion, four hundred sixty-six million, nine hundred ninety-seven thousand, four hundred thirty-three".
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