Base | Representation |
---|---|
bin | 1100000010001001110110… |
… | …0001101001100001001011 |
3 | 1201211212220001011120221111 |
4 | 3000202131201221201023 |
5 | 3213234334340143103 |
6 | 44050145222151151 |
7 | 2533630015225642 |
oct | 300423541514113 |
9 | 51755801146844 |
10 | 13231142115403 |
11 | 4241328875153 |
12 | 159834a8824b7 |
13 | 74c8cab779c4 |
14 | 33a5680d4359 |
15 | 17e28bb3986d |
hex | c089d86984b |
13231142115403 has 4 divisors (see below), whose sum is σ = 13231150261560. Its totient is φ = 13231133969248.
The previous prime is 13231142115337. The next prime is 13231142115419. The reversal of 13231142115403 is 30451124113231.
It is a happy number.
It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13231142115403 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (13231142155403) 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, 712525 + ... + 5193262.
It is an arithmetic number, because the mean of its divisors is an integer number (3307787565390).
Almost surely, 213231142115403 is an apocalyptic number.
13231142115403 is a deficient number, since it is larger than the sum of its proper divisors (8146157).
13231142115403 is an equidigital number, since it uses as much as digits as its factorization.
13231142115403 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 8146156.
The product of its (nonzero) digits is 8640, while the sum is 31.
Adding to 13231142115403 its reverse (30451124113231), we get a palindrome (43682266228634).
The spelling of 13231142115403 in words is "thirteen trillion, two hundred thirty-one billion, one hundred forty-two million, one hundred fifteen thousand, four hundred three".
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