Base | Representation |
---|---|
bin | 11110011111011110010111… |
… | …001011110010100011111011 |
3 | 122120211020122021011220020122 |
4 | 132133132113023302203323 |
5 | 120034131101341131212 |
6 | 1153114355354043455 |
7 | 40150500402005153 |
oct | 3637362713624373 |
9 | 576736567156218 |
10 | 134104300333307 |
11 | 3980335049aa67 |
12 | 1305a38aa1958b |
13 | 59a9c9aa98475 |
14 | 251897b731763 |
15 | 107856208b272 |
hex | 79f7972f28fb |
134104300333307 has 2 divisors, whose sum is σ = 134104300333308. Its totient is φ = 134104300333306.
The previous prime is 134104300333247. The next prime is 134104300333309. The reversal of 134104300333307 is 703333003401431.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 134104300333307 - 210 = 134104300332283 is a prime.
Together with 134104300333309, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (134104300333309) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (31) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67052150166653 + 67052150166654.
It is an arithmetic number, because the mean of its divisors is an integer number (67052150166654).
Almost surely, 2134104300333307 is an apocalyptic number.
134104300333307 is a deficient number, since it is larger than the sum of its proper divisors (1).
134104300333307 is an equidigital number, since it uses as much as digits as its factorization.
134104300333307 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 81648, while the sum is 35.
Adding to 134104300333307 its reverse (703333003401431), we get a palindrome (837437303734738).
The spelling of 134104300333307 in words is "one hundred thirty-four trillion, one hundred four billion, three hundred million, three hundred thirty-three thousand, three hundred seven".
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