Base | Representation |
---|---|
bin | 11101100011111011001101… |
… | …101000100100011011111111 |
3 | 122001100000222111010202020112 |
4 | 131203323031220210123333 |
5 | 114020104300130423312 |
6 | 1140302423325340235 |
7 | 36246031200136361 |
oct | 3543731550443377 |
9 | 561300874122215 |
10 | 130012110014207 |
11 | 3847590a770133 |
12 | 126b927470767b |
13 | 5771121458675 |
14 | 2416896656d31 |
15 | 1006da77ead22 |
hex | 763ecda246ff |
130012110014207 has 2 divisors, whose sum is σ = 130012110014208. Its totient is φ = 130012110014206.
The previous prime is 130012110014057. The next prime is 130012110014239. The reversal of 130012110014207 is 702410011210031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 130012110014207 - 210 = 130012110013183 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130012110514207) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65006055007103 + 65006055007104.
It is an arithmetic number, because the mean of its divisors is an integer number (65006055007104).
Almost surely, 2130012110014207 is an apocalyptic number.
130012110014207 is a deficient number, since it is larger than the sum of its proper divisors (1).
130012110014207 is an equidigital number, since it uses as much as digits as its factorization.
130012110014207 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 336, while the sum is 23.
Adding to 130012110014207 its reverse (702410011210031), we get a palindrome (832422121224238).
The spelling of 130012110014207 in words is "one hundred thirty trillion, twelve billion, one hundred ten million, fourteen thousand, two hundred seven".
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