Base | Representation |
---|---|
bin | 100110101001110010111… |
… | …0000110000100011110111 |
3 | 200210212022200110011220111 |
4 | 1031103211300300203313 |
5 | 1144014331142422020 |
6 | 15144255412443451 |
7 | 1055545034565034 |
oct | 115234560604367 |
9 | 20725280404814 |
10 | 5312434342135 |
11 | 1768a99767078 |
12 | 719703619587 |
13 | 2c6c6403146b |
14 | 1451a208a38b |
15 | 932c6849d5a |
hex | 4d4e5c308f7 |
5312434342135 has 32 divisors (see below), whose sum is σ = 6435843858432. Its totient is φ = 4209513235200.
The previous prime is 5312434342121. The next prime is 5312434342147.
5312434342135 is nontrivially palindromic in base 10.
5312434342135 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 5312434342135 - 215 = 5312434309367 is a prime.
It is a Duffinian number.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 429212067 + ... + 429224443.
It is an arithmetic number, because the mean of its divisors is an integer number (201120120576).
Almost surely, 25312434342135 is an apocalyptic number.
5312434342135 is a deficient number, since it is larger than the sum of its proper divisors (1123409516297).
5312434342135 is a wasteful number, since it uses less digits than its factorization.
5312434342135 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 14143.
The product of its digits is 518400, while the sum is 40.
The spelling of 5312434342135 in words is "five trillion, three hundred twelve billion, four hundred thirty-four million, three hundred forty-two thousand, one hundred thirty-five".
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