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13122633601219 is a prime number
BaseRepresentation
bin1011111011110101100111…
…1010100001000011000011
31201110111210220020221110211
42332331121322201003003
53210000123200214334
643524242112134551
72523036044122432
oct276753172410303
951414726227424
1013122633601219
1141aa30661a44a
12157b307537457
137425c767b8ac
143351d3039119
1517b53a958a64
hexbef59ea10c3

13122633601219 has 2 divisors, whose sum is σ = 13122633601220. Its totient is φ = 13122633601218.

The previous prime is 13122633601193. The next prime is 13122633601229. The reversal of 13122633601219 is 91210633622131.

It is a strong prime.

It is an emirp because it is prime and its reverse (91210633622131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13122633601219 - 239 = 12572877787331 is a prime.

It is a super-3 number, since 3×131226336012193 (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is not a weakly prime, because it can be changed into another prime (13122633601229) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6561316800609 + 6561316800610.

It is an arithmetic number, because the mean of its divisors is an integer number (6561316800610).

Almost surely, 213122633601219 is an apocalyptic number.

13122633601219 is a deficient number, since it is larger than the sum of its proper divisors (1).

13122633601219 is an equidigital number, since it uses as much as digits as its factorization.

13122633601219 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 69984, while the sum is 40.

The spelling of 13122633601219 in words is "thirteen trillion, one hundred twenty-two billion, six hundred thirty-three million, six hundred one thousand, two hundred nineteen".