Base | Representation |
---|---|
bin | 1010111010101011000010… |
… | …1100101100001110100101 |
3 | 1120111111002220011110121221 |
4 | 2232222300230230032211 |
5 | 3033124323400134113 |
6 | 41310052140540341 |
7 | 2346124041126214 |
oct | 256526054541645 |
9 | 46444086143557 |
10 | 12003103130533 |
11 | 390853a884799 |
12 | 141a346b516b1 |
13 | 690b705727b8 |
14 | 2d6d4bb8d07b |
15 | 15c36558698d |
hex | aeab0b2c3a5 |
12003103130533 has 2 divisors, whose sum is σ = 12003103130534. Its totient is φ = 12003103130532.
The previous prime is 12003103130507. The next prime is 12003103130537. The reversal of 12003103130533 is 33503130130021.
12003103130533 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 12000522287329 + 2580843204 = 3464177^2 + 50802^2 .
It is a cyclic number.
It is not a de Polignac number, because 12003103130533 - 213 = 12003103122341 is a prime.
It is a super-2 number, since 2×120031031305332 (a number of 27 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 12003103130498 and 12003103130507.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12003103130537) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6001551565266 + 6001551565267.
It is an arithmetic number, because the mean of its divisors is an integer number (6001551565267).
Almost surely, 212003103130533 is an apocalyptic number.
It is an amenable number.
12003103130533 is a deficient number, since it is larger than the sum of its proper divisors (1).
12003103130533 is an equidigital number, since it uses as much as digits as its factorization.
12003103130533 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2430, while the sum is 25.
Adding to 12003103130533 its reverse (33503130130021), we get a palindrome (45506233260554).
The spelling of 12003103130533 in words is "twelve trillion, three billion, one hundred three million, one hundred thirty thousand, five hundred thirty-three".
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