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131312312200019 is a prime number
BaseRepresentation
bin11101110110110110000111…
…101110010010101101010011
3122012221100000210221022021002
4131312312013232102231103
5114202410103410400034
61143140013214452215
736442000336311056
oct3566660756225523
9565840023838232
10131312312200019
1138927269a28535
121288925780406b
13583690c3c3c82
14245d79a9d2a9d
15102ab0435ce7e
hex776d87b92b53

131312312200019 has 2 divisors, whose sum is σ = 131312312200020. Its totient is φ = 131312312200018.

The previous prime is 131312312199977. The next prime is 131312312200033. The reversal of 131312312200019 is 910002213213131.

It is a strong prime.

It is an emirp because it is prime and its reverse (910002213213131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131312312200019 - 28 = 131312312199763 is a prime.

It is a super-2 number, since 2×1313123122000192 (a number of 29 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 131312312199964 and 131312312200000.

It is not a weakly prime, because it can be changed into another prime (131312312200519) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65656156100009 + 65656156100010.

It is an arithmetic number, because the mean of its divisors is an integer number (65656156100010).

Almost surely, 2131312312200019 is an apocalyptic number.

131312312200019 is a deficient number, since it is larger than the sum of its proper divisors (1).

131312312200019 is an equidigital number, since it uses as much as digits as its factorization.

131312312200019 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1944, while the sum is 29.

The spelling of 131312312200019 in words is "one hundred thirty-one trillion, three hundred twelve billion, three hundred twelve million, two hundred thousand, nineteen".