Base | Representation |
---|---|
bin | 11101100101010110100100… |
… | …000100011111110010000011 |
3 | 122001200102011022222211220112 |
4 | 131211112210010133302003 |
5 | 114023211140404034112 |
6 | 1140415444352501535 |
7 | 36256102654222652 |
oct | 3545264404376203 |
9 | 561612138884815 |
10 | 130110196939907 |
11 | 38503474230198 |
12 | 1271428984a8ab |
13 | 577a453807582 |
14 | 241b51d5a9199 |
15 | 10096e8ae3a22 |
hex | 7655a411fc83 |
130110196939907 has 2 divisors, whose sum is σ = 130110196939908. Its totient is φ = 130110196939906.
The previous prime is 130110196939873. The next prime is 130110196939909. The reversal of 130110196939907 is 709939691011031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 130110196939907 - 212 = 130110196935811 is a prime.
Together with 130110196939909, 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 (130110196939909) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65055098469953 + 65055098469954.
It is an arithmetic number, because the mean of its divisors is an integer number (65055098469954).
Almost surely, 2130110196939907 is an apocalyptic number.
130110196939907 is a deficient number, since it is larger than the sum of its proper divisors (1).
130110196939907 is an equidigital number, since it uses as much as digits as its factorization.
130110196939907 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2480058, while the sum is 59.
The spelling of 130110196939907 in words is "one hundred thirty trillion, one hundred ten billion, one hundred ninety-six million, nine hundred thirty-nine thousand, nine hundred seven".
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