Base | Representation |
---|---|
bin | 1011001100001001110011… |
… | …1111001000001011010011 |
3 | 1121120012021111120201220011 |
4 | 2303002130333020023103 |
5 | 3103034401023130003 |
6 | 42100040240255351 |
7 | 2406615135402352 |
oct | 263023477101323 |
9 | 47505244521804 |
10 | 12303420130003 |
11 | 3a1393aa06422 |
12 | 146859a611557 |
13 | 6b3290c71952 |
14 | 3076bb04b999 |
15 | 1650909b146d |
hex | b309cfc82d3 |
12303420130003 has 2 divisors, whose sum is σ = 12303420130004. Its totient is φ = 12303420130002.
The previous prime is 12303420129971. The next prime is 12303420130073. The reversal of 12303420130003 is 30003102430321.
12303420130003 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12303420130003 - 25 = 12303420129971 is a prime.
It is a super-3 number, since 3×123034201300033 (a number of 40 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (12303420130073) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6151710065001 + 6151710065002.
It is an arithmetic number, because the mean of its divisors is an integer number (6151710065002).
Almost surely, 212303420130003 is an apocalyptic number.
12303420130003 is a deficient number, since it is larger than the sum of its proper divisors (1).
12303420130003 is an equidigital number, since it uses as much as digits as its factorization.
12303420130003 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1296, while the sum is 22.
Adding to 12303420130003 its reverse (30003102430321), we get a palindrome (42306522560324).
The spelling of 12303420130003 in words is "twelve trillion, three hundred three billion, four hundred twenty million, one hundred thirty thousand, three".
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