Base | Representation |
---|---|
bin | 100101001100001001100… |
… | …1000010111000101011111 |
3 | 200002122020111222210121011 |
4 | 1022120103020113011133 |
5 | 1132221002000303210 |
6 | 14512040304320051 |
7 | 1035165400332202 |
oct | 112302310270537 |
9 | 20078214883534 |
10 | 5111332041055 |
11 | 16a0781759849 |
12 | 6a673a913027 |
13 | 2b0cc5537136 |
14 | 13956582b939 |
15 | 8ce56737d8a |
hex | 4a61321715f |
5111332041055 has 8 divisors (see below), whose sum is σ = 6456419420400. Its totient is φ = 3873851652096.
The previous prime is 5111332041019. The next prime is 5111332041059. The reversal of 5111332041055 is 5501402331115.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 5111332041055 - 211 = 5111332039007 is a prime.
It is a super-2 number, since 2×51113320410552 (a number of 26 digits) contains 22 as substring.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (5111332041059) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 26901747490 + ... + 26901747679.
It is an arithmetic number, because the mean of its divisors is an integer number (807052427550).
Almost surely, 25111332041055 is an apocalyptic number.
5111332041055 is a deficient number, since it is larger than the sum of its proper divisors (1345087379345).
5111332041055 is a wasteful number, since it uses less digits than its factorization.
5111332041055 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 53803495193.
The product of its (nonzero) digits is 9000, while the sum is 31.
The spelling of 5111332041055 in words is "five trillion, one hundred eleven billion, three hundred thirty-two million, forty-one thousand, fifty-five".
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