Base | Representation |
---|---|
bin | 100111100101101010110… |
… | …1110011010100111101011 |
3 | 201021011012122100101120111 |
4 | 1033023111232122213223 |
5 | 1203121144130431103 |
6 | 15323322323435151 |
7 | 1101046256362066 |
oct | 117132556324753 |
9 | 21234178311514 |
10 | 5441014311403 |
11 | 1808580806090 |
12 | 73a608825ab7 |
13 | 306115963108 |
14 | 14b4bcb0b8dd |
15 | 967eebad86d |
hex | 4f2d5b9a9eb |
5441014311403 has 4 divisors (see below), whose sum is σ = 5935651976088. Its totient is φ = 4946376646720.
The previous prime is 5441014311389. The next prime is 5441014311419. The reversal of 5441014311403 is 3041134101445.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 5441014311403 - 213 = 5441014303211 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (5441014311433) 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, 247318832326 + ... + 247318832347.
It is an arithmetic number, because the mean of its divisors is an integer number (1483912994022).
Almost surely, 25441014311403 is an apocalyptic number.
5441014311403 is a deficient number, since it is larger than the sum of its proper divisors (494637664685).
5441014311403 is a wasteful number, since it uses less digits than its factorization.
5441014311403 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 494637664684.
The product of its (nonzero) digits is 11520, while the sum is 31.
Adding to 5441014311403 its reverse (3041134101445), we get a palindrome (8482148412848).
The spelling of 5441014311403 in words is "five trillion, four hundred forty-one billion, fourteen million, three hundred eleven thousand, four hundred three".
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