Base | Representation |
---|---|
bin | 1011000001001000001101… |
… | …1000011100111100000011 |
3 | 1120220002101022202210112020 |
4 | 2300102003120130330003 |
5 | 3041433444033000003 |
6 | 41433033354532523 |
7 | 2360131340015664 |
oct | 260220330347403 |
9 | 46802338683466 |
10 | 12114012000003 |
11 | 3950583a649a1 |
12 | 143793a137143 |
13 | 69b467150194 |
14 | 2dc46d95b36b |
15 | 1601a735dd53 |
hex | b048361cf03 |
12114012000003 has 16 divisors (see below), whose sum is σ = 16460380154016. Its totient is φ = 7921892759040.
The previous prime is 12114012000001. The next prime is 12114012000011. The reversal of 12114012000003 is 30000021041121.
It is not a de Polignac number, because 12114012000003 - 21 = 12114012000001 is a prime.
It is a super-3 number, since 3×121140120000033 (a number of 40 digits) contains 333 as substring.
It is not an unprimeable number, because it can be changed into a prime (12114012000001) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 7627000 + ... + 9077397.
It is an arithmetic number, because the mean of its divisors is an integer number (1028773759626).
Almost surely, 212114012000003 is an apocalyptic number.
12114012000003 is a deficient number, since it is larger than the sum of its proper divisors (4346368154013).
12114012000003 is a wasteful number, since it uses less digits than its factorization.
12114012000003 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 16709014.
The product of its (nonzero) digits is 48, while the sum is 15.
Adding to 12114012000003 its reverse (30000021041121), we get a palindrome (42114033041124).
The spelling of 12114012000003 in words is "twelve trillion, one hundred fourteen billion, twelve million, three", and thus it is an aban number.
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