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130011040231049 is a prime number
BaseRepresentation
bin11101100011111010001101…
…110111101010111010001001
3122001022221012221011010122002
4131203322031313222322021
5114020100102244343144
61140302125232240345
736245662530134135
oct3543721567527211
9561287187133562
10130011040231049
113847541091421a
12126b90163a00b5
135770cb0927593
1424167d45423c5
151006d4393314e
hex763e8ddeae89

130011040231049 has 2 divisors, whose sum is σ = 130011040231050. Its totient is φ = 130011040231048.

The previous prime is 130011040231021. The next prime is 130011040231099. The reversal of 130011040231049 is 940132040110031.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 66716795761225 + 63294244469824 = 8168035^2 + 7955768^2 .

It is an emirp because it is prime and its reverse (940132040110031) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 130011040231049 - 28 = 130011040230793 is a prime.

It is not a weakly prime, because it can be changed into another prime (130011040231099) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65005520115524 + 65005520115525.

It is an arithmetic number, because the mean of its divisors is an integer number (65005520115525).

Almost surely, 2130011040231049 is an apocalyptic number.

It is an amenable number.

130011040231049 is a deficient number, since it is larger than the sum of its proper divisors (1).

130011040231049 is an equidigital number, since it uses as much as digits as its factorization.

130011040231049 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 2592, while the sum is 29.

The spelling of 130011040231049 in words is "one hundred thirty trillion, eleven billion, forty million, two hundred thirty-one thousand, forty-nine".