Base | Representation |
---|---|
bin | 11000011110111011… |
… | …11100111101100011 |
3 | 1020221001112022111012 |
4 | 30033131330331203 |
5 | 203404432141003 |
6 | 10012150211135 |
7 | 643505441612 |
oct | 141735747543 |
9 | 36831468435 |
10 | 13144412003 |
11 | 5635750318 |
12 | 266a053aab |
13 | 13162952a8 |
14 | 8c99db879 |
15 | 51de7a8d8 |
hex | 30f77cf63 |
13144412003 has 4 divisors (see below), whose sum is σ = 13144935888. Its totient is φ = 13143888120.
The previous prime is 13144411961. The next prime is 13144412009. The reversal of 13144412003 is 30021444131.
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 13144412003 - 214 = 13144395619 is a prime.
It is a super-2 number, since 2×131444120032 (a number of 21 digits) contains 22 as substring.
It is a Duffinian number.
It is a self number, because there is not a number n which added to its sum of digits gives 13144412003.
It is not an unprimeable number, because it can be changed into a prime (13144412009) 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, 222308 + ... + 275153.
It is an arithmetic number, because the mean of its divisors is an integer number (3286233972).
Almost surely, 213144412003 is an apocalyptic number.
13144412003 is a deficient number, since it is larger than the sum of its proper divisors (523885).
13144412003 is an equidigital number, since it uses as much as digits as its factorization.
13144412003 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 523884.
The product of its (nonzero) digits is 1152, while the sum is 23.
Adding to 13144412003 its reverse (30021444131), we get a palindrome (43165856134).
The spelling of 13144412003 in words is "thirteen billion, one hundred forty-four million, four hundred twelve thousand, three".
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