Base | Representation |
---|---|
bin | 1100001101111111001100… |
… | …0011001101101001010011 |
3 | 1202120022122020211112202002 |
4 | 3003133303003031221103 |
5 | 3230102213223043103 |
6 | 44323410254544215 |
7 | 2554414646410553 |
oct | 303376303155123 |
9 | 52508566745662 |
10 | 13434440440403 |
11 | 430a572436784 |
12 | 160b826b0806b |
13 | 765b29610978 |
14 | 346332277163 |
15 | 1846d9930488 |
hex | c37f30cda53 |
13434440440403 has 4 divisors (see below), whose sum is σ = 13434491166720. Its totient is φ = 13434389714088.
The previous prime is 13434440440339. The next prime is 13434440440411. The reversal of 13434440440403 is 30404404443431.
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 13434440440403 - 26 = 13434440440339 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (13434440440493) 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, 24963800 + ... + 25496277.
It is an arithmetic number, because the mean of its divisors is an integer number (3358622791680).
Almost surely, 213434440440403 is an apocalyptic number.
13434440440403 is a deficient number, since it is larger than the sum of its proper divisors (50726317).
13434440440403 is an equidigital number, since it uses as much as digits as its factorization.
13434440440403 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 50726316.
The product of its (nonzero) digits is 442368, while the sum is 38.
Adding to 13434440440403 its reverse (30404404443431), we get a palindrome (43838844883834).
The spelling of 13434440440403 in words is "thirteen trillion, four hundred thirty-four billion, four hundred forty million, four hundred forty thousand, four hundred three".
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