Base | Representation |
---|---|
bin | 1011111010100010001101… |
… | …1010100010111000110001 |
3 | 1201101101000120001101111121 |
4 | 2332220203122202320301 |
5 | 3204113300032222001 |
6 | 43510100314142241 |
7 | 2521314161145223 |
oct | 276504332427061 |
9 | 51341016041447 |
10 | 13100244414001 |
11 | 41a08674a3738 |
12 | 1576ab9421981 |
13 | 74046a002b14 |
14 | 3340ab70a413 |
15 | 17ab7a0a06a1 |
hex | bea236a2e31 |
13100244414001 has 8 divisors (see below), whose sum is σ = 13669878386304. Its totient is φ = 12530615285760.
The previous prime is 13100244413953. The next prime is 13100244414007. The reversal of 13100244414001 is 10041444200131.
13100244414001 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
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 13100244414001 - 215 = 13100244381233 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (13100244414007) 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, 4991281 + ... + 7149361.
It is an arithmetic number, because the mean of its divisors is an integer number (1708734798288).
Almost surely, 213100244414001 is an apocalyptic number.
It is an amenable number.
13100244414001 is a deficient number, since it is larger than the sum of its proper divisors (569633972303).
13100244414001 is a wasteful number, since it uses less digits than its factorization.
13100244414001 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2422031.
The product of its (nonzero) digits is 1536, while the sum is 25.
Adding to 13100244414001 its reverse (10041444200131), we get a palindrome (23141688614132).
The spelling of 13100244414001 in words is "thirteen trillion, one hundred billion, two hundred forty-four million, four hundred fourteen thousand, one".
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