Base | Representation |
---|---|
bin | 11101100100000010101011… |
… | …101011001100100000111111 |
3 | 122001100210200102120000011122 |
4 | 131210002223223030200333 |
5 | 114020222211313240103 |
6 | 1140310231230141155 |
7 | 36246434014531061 |
oct | 3544025353144077 |
9 | 561323612500148 |
10 | 130020130211903 |
11 | 38479255983398 |
12 | 126ba9326517bb |
13 | 5771acbc20b89 |
14 | 241701789b731 |
15 | 10071c6985b38 |
hex | 7640abacc83f |
130020130211903 has 2 divisors, whose sum is σ = 130020130211904. Its totient is φ = 130020130211902.
The previous prime is 130020130211879. The next prime is 130020130211987. The reversal of 130020130211903 is 309112031020031.
It is a weak prime.
It is an emirp because it is prime and its reverse (309112031020031) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 130020130211903 - 214 = 130020130195519 is a prime.
It is a super-2 number, since 2×1300201302119032 (a number of 29 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (130020130291903) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65010065105951 + 65010065105952.
It is an arithmetic number, because the mean of its divisors is an integer number (65010065105952).
Almost surely, 2130020130211903 is an apocalyptic number.
130020130211903 is a deficient number, since it is larger than the sum of its proper divisors (1).
130020130211903 is an equidigital number, since it uses as much as digits as its factorization.
130020130211903 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 972, while the sum is 26.
Adding to 130020130211903 its reverse (309112031020031), we get a palindrome (439132161231934).
The spelling of 130020130211903 in words is "one hundred thirty trillion, twenty billion, one hundred thirty million, two hundred eleven thousand, nine hundred three".
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