Base | Representation |
---|---|
bin | 11101110010011101111000… |
… | …100011011111100100000011 |
3 | 122011212120100102100202221022 |
4 | 131302131320203133210003 |
5 | 114132442331430001011 |
6 | 1142345451052220055 |
7 | 36411155604645143 |
oct | 3562357043374403 |
9 | 564776312322838 |
10 | 131011410000131 |
11 | 388206995370a1 |
12 | 1283aa8016a62b |
13 | 58144276ab59b |
14 | 244cdb3b20923 |
15 | 1022d9288a8db |
hex | 7727788df903 |
131011410000131 has 2 divisors, whose sum is σ = 131011410000132. Its totient is φ = 131011410000130.
The previous prime is 131011410000109. The next prime is 131011410000181. The reversal of 131011410000131 is 131000014110131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 131011410000131 - 26 = 131011410000067 is a prime.
It is a super-2 number, since 2×1310114100001312 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (131011410000181) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65505705000065 + 65505705000066.
It is an arithmetic number, because the mean of its divisors is an integer number (65505705000066).
Almost surely, 2131011410000131 is an apocalyptic number.
131011410000131 is a deficient number, since it is larger than the sum of its proper divisors (1).
131011410000131 is an equidigital number, since it uses as much as digits as its factorization.
131011410000131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 36, while the sum is 17.
Adding to 131011410000131 its reverse (131000014110131), we get a palindrome (262011424110262).
The spelling of 131011410000131 in words is "one hundred thirty-one trillion, eleven billion, four hundred ten million, one hundred thirty-one", and thus it is an aban number.
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