Base | Representation |
---|---|
bin | 1011111011110100011111… |
… | …1001011100011100000011 |
3 | 1201110111000211110222201212 |
4 | 2332331013321130130003 |
5 | 3204444013310032212 |
6 | 43524152113320335 |
7 | 2523025413634223 |
oct | 276750771343403 |
9 | 51414024428655 |
10 | 13122331330307 |
11 | 41aa170a34881 |
12 | 157b2422660ab |
13 | 742579b67274 |
14 | 3351a4c34083 |
15 | 17b51e14be22 |
hex | bef47e5c703 |
13122331330307 has 2 divisors, whose sum is σ = 13122331330308. Its totient is φ = 13122331330306.
The previous prime is 13122331330261. The next prime is 13122331330357. The reversal of 13122331330307 is 70303313322131.
It is a weak prime.
It is an emirp because it is prime and its reverse (70303313322131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13122331330307 - 222 = 13122327136003 is a prime.
It is a super-3 number, since 3×131223313303073 (a number of 40 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (13122331330357) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6561165665153 + 6561165665154.
It is an arithmetic number, because the mean of its divisors is an integer number (6561165665154).
Almost surely, 213122331330307 is an apocalyptic number.
13122331330307 is a deficient number, since it is larger than the sum of its proper divisors (1).
13122331330307 is an equidigital number, since it uses as much as digits as its factorization.
13122331330307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 20412, while the sum is 32.
Adding to 13122331330307 its reverse (70303313322131), we get a palindrome (83425644652438).
The spelling of 13122331330307 in words is "thirteen trillion, one hundred twenty-two billion, three hundred thirty-one million, three hundred thirty thousand, three hundred seven".
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