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13001312120999 is a prime number
BaseRepresentation
bin1011110100110001101010…
…0101110011000010100111
31201000220122210111011220112
42331030122211303002213
53201003141400332444
643352415331523235
72511212430424136
oct275143245630247
951026583434815
1013001312120999
114162909826714
12155b8a915951b
13734031778487
1432d3a43bd91d
151782d996d69e
hexbd31a9730a7

13001312120999 has 2 divisors, whose sum is σ = 13001312121000. Its totient is φ = 13001312120998.

The previous prime is 13001312120939. The next prime is 13001312121073. The reversal of 13001312120999 is 99902121310031.

It is a weak prime.

It is an emirp because it is prime and its reverse (99902121310031) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13001312120999 - 28 = 13001312120743 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13001312120939) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6500656060499 + 6500656060500.

It is an arithmetic number, because the mean of its divisors is an integer number (6500656060500).

Almost surely, 213001312120999 is an apocalyptic number.

13001312120999 is a deficient number, since it is larger than the sum of its proper divisors (1).

13001312120999 is an equidigital number, since it uses as much as digits as its factorization.

13001312120999 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 26244, while the sum is 41.

The spelling of 13001312120999 in words is "thirteen trillion, one billion, three hundred twelve million, one hundred twenty thousand, nine hundred ninety-nine".