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14130452113 = 831279133109
BaseRepresentation
bin11010010100011110…
…11001011010010001
31100110202221100020021
431022033121122101
5212414343431423
610254052343441
71010110601251
oct151217313221
940422840207
1014130452113
115aa12a98a8
1228a4314b81
13144265538b
14980953961
1557a80095d
hex34a3d9691

14130452113 has 8 divisors (see below), whose sum is σ = 14311987200. Its totient is φ = 13949185968.

The previous prime is 14130452081. The next prime is 14130452117. The reversal of 14130452113 is 31125403141.

14130452113 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 14130452113 - 25 = 14130452081 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (14130452117) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 39603 + ... + 172711.

It is an arithmetic number, because the mean of its divisors is an integer number (1788998400).

Almost surely, 214130452113 is an apocalyptic number.

It is an amenable number.

14130452113 is a deficient number, since it is larger than the sum of its proper divisors (181535087).

14130452113 is a wasteful number, since it uses less digits than its factorization.

14130452113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 134471.

The product of its (nonzero) digits is 1440, while the sum is 25.

Adding to 14130452113 its reverse (31125403141), we get a palindrome (45255855254).

The spelling of 14130452113 in words is "fourteen billion, one hundred thirty million, four hundred fifty-two thousand, one hundred thirteen".

Divisors: 1 83 1279 106157 133109 11048047 170246411 14130452113