Base | Representation |
---|---|
bin | 1010111010101011000101… |
… | …0111110101010111000011 |
3 | 1120111111010200110001110221 |
4 | 2232222301113311113003 |
5 | 3033124334244041011 |
6 | 41310053223203511 |
7 | 2346124236221623 |
oct | 256526127652703 |
9 | 46444120401427 |
10 | 12003114440131 |
11 | 39085461a9871 |
12 | 141a34a8a6597 |
13 | 690b72a02479 |
14 | 2d6d4d494883 |
15 | 15c36656c971 |
hex | aeab15f55c3 |
12003114440131 has 32 divisors (see below), whose sum is σ = 12571875634176. Its totient is φ = 11452163184000.
The previous prime is 12003114440099. The next prime is 12003114440147. The reversal of 12003114440131 is 13104441130021.
It is a cyclic number.
It is not a de Polignac number, because 12003114440131 - 25 = 12003114440099 is a prime.
It is a super-2 number, since 2×120031144401312 (a number of 27 digits) contains 22 as substring.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 12003114440096 and 12003114440105.
It is not an unprimeable number, because it can be changed into a prime (12003114444131) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 238790541 + ... + 238840801.
It is an arithmetic number, because the mean of its divisors is an integer number (392871113568).
Almost surely, 212003114440131 is an apocalyptic number.
12003114440131 is a deficient number, since it is larger than the sum of its proper divisors (568761194045).
12003114440131 is an equidigital number, since it uses as much as digits as its factorization.
12003114440131 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 51283.
The product of its (nonzero) digits is 1152, while the sum is 25.
Adding to 12003114440131 its reverse (13104441130021), we get a palindrome (25107555570152).
The spelling of 12003114440131 in words is "twelve trillion, three billion, one hundred fourteen million, four hundred forty thousand, one hundred thirty-one".
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