Base | Representation |
---|---|
bin | 11001001101010000100101… |
… | …01011000100001010101111 |
3 | 21021021011111221010210200112 |
4 | 30212220102223010022233 |
5 | 24231141004231420302 |
6 | 313520420210450235 |
7 | 14450524564055612 |
oct | 1446502253041257 |
9 | 237234457123615 |
10 | 55431161201327 |
11 | 1673125045779a |
12 | 6272b1b89397b |
13 | 24c119cc49032 |
14 | d98c4d8ac179 |
15 | 661d597c8552 |
hex | 326a12ac42af |
55431161201327 has 2 divisors, whose sum is σ = 55431161201328. Its totient is φ = 55431161201326.
The previous prime is 55431161201323. The next prime is 55431161201329. The reversal of 55431161201327 is 72310216113455.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 55431161201327 - 22 = 55431161201323 is a prime.
It is a super-2 number, since 2×554311612013272 (a number of 28 digits) contains 22 as substring.
Together with 55431161201329, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (55431161201323) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27715580600663 + 27715580600664.
It is an arithmetic number, because the mean of its divisors is an integer number (27715580600664).
Almost surely, 255431161201327 is an apocalyptic number.
55431161201327 is a deficient number, since it is larger than the sum of its proper divisors (1).
55431161201327 is an equidigital number, since it uses as much as digits as its factorization.
55431161201327 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 151200, while the sum is 41.
The spelling of 55431161201327 in words is "fifty-five trillion, four hundred thirty-one billion, one hundred sixty-one million, two hundred one thousand, three hundred twenty-seven".
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