Base | Representation |
---|---|
bin | 1011111011010110101100… |
… | …1110000110101101011111 |
3 | 1201102201101111010020120211 |
4 | 2332311223032012231133 |
5 | 3204331124001424203 |
6 | 43520350351153251 |
7 | 2522323264603144 |
oct | 276655316065537 |
9 | 51381344106524 |
10 | 13114334014303 |
11 | 41a683772aa05 |
12 | 157978bb15827 |
13 | 7418a4053a12 |
14 | 334a46a683cb |
15 | 17b201ed1c6d |
hex | bed6b386b5f |
13114334014303 has 4 divisors (see below), whose sum is σ = 13115859122800. Its totient is φ = 13112808905808.
The previous prime is 13114334014277. The next prime is 13114334014337. The reversal of 13114334014303 is 30341043341131.
It is a happy number.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13114334014303 is a prime.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (13114334014403) 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, 762541350 + ... + 762558547.
It is an arithmetic number, because the mean of its divisors is an integer number (3278964780700).
Almost surely, 213114334014303 is an apocalyptic number.
13114334014303 is a deficient number, since it is larger than the sum of its proper divisors (1525108497).
13114334014303 is an equidigital number, since it uses as much as digits as its factorization.
13114334014303 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1525108496.
The product of its (nonzero) digits is 15552, while the sum is 31.
Adding to 13114334014303 its reverse (30341043341131), we get a palindrome (43455377355434).
The spelling of 13114334014303 in words is "thirteen trillion, one hundred fourteen billion, three hundred thirty-four million, fourteen thousand, three hundred three".
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