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131404302232019 is a prime number
BaseRepresentation
bin11101111000001011110010…
…110000011010100111010011
3122020021010110202102010002022
4131320023302300122213103
5114210412002332411034
61143250145244242055
736451435051552145
oct3570136260324723
9566233422363068
10131404302232019
11389622849a8073
12128a304ab4032b
1358424ac5b047b
142464005d12c95
15102d1da27db2e
hex7782f2c1a9d3

131404302232019 has 2 divisors, whose sum is σ = 131404302232020. Its totient is φ = 131404302232018.

The previous prime is 131404302231919. The next prime is 131404302232079. The reversal of 131404302232019 is 910232203404131.

It is a strong prime.

It is an emirp because it is prime and its reverse (910232203404131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131404302232019 - 212 = 131404302227923 is a prime.

It is a super-2 number, since 2×1314043022320192 (a number of 29 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 131404302232019.

It is not a weakly prime, because it can be changed into another prime (131404302232079) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65702151116009 + 65702151116010.

It is an arithmetic number, because the mean of its divisors is an integer number (65702151116010).

Almost surely, 2131404302232019 is an apocalyptic number.

131404302232019 is a deficient number, since it is larger than the sum of its proper divisors (1).

131404302232019 is an equidigital number, since it uses as much as digits as its factorization.

131404302232019 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 31104, while the sum is 35.

The spelling of 131404302232019 in words is "one hundred thirty-one trillion, four hundred four billion, three hundred two million, two hundred thirty-two thousand, nineteen".