Base | Representation |
---|---|
bin | 1011001100101110100011… |
… | …0111111101101000010111 |
3 | 1121121010201212210101112211 |
4 | 2303023220313331220113 |
5 | 3103220101003123212 |
6 | 42104351053254251 |
7 | 2410414431520303 |
oct | 263135067755027 |
9 | 47533655711484 |
10 | 12313283254807 |
11 | 3a18041433037 |
12 | 146a491794387 |
13 | 6b41a34a7182 |
14 | 307d74d40d03 |
15 | 16546b8339a7 |
hex | b32e8dfda17 |
12313283254807 has 2 divisors, whose sum is σ = 12313283254808. Its totient is φ = 12313283254806.
The previous prime is 12313283254739. The next prime is 12313283254819. The reversal of 12313283254807 is 70845238231321.
It is a strong prime.
It is an emirp because it is prime and its reverse (70845238231321) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 12313283254807 - 211 = 12313283252759 is a prime.
It is a super-2 number, since 2×123132832548072 (a number of 27 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12313283255807) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6156641627403 + 6156641627404.
It is an arithmetic number, because the mean of its divisors is an integer number (6156641627404).
Almost surely, 212313283254807 is an apocalyptic number.
12313283254807 is a deficient number, since it is larger than the sum of its proper divisors (1).
12313283254807 is an equidigital number, since it uses as much as digits as its factorization.
12313283254807 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1935360, while the sum is 49.
The spelling of 12313283254807 in words is "twelve trillion, three hundred thirteen billion, two hundred eighty-three million, two hundred fifty-four thousand, eight hundred seven".
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