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11004930132017 is a prime number
BaseRepresentation
bin1010000000100100100011…
…1100111000100000110001
31102222001122111120100000112
42200021020330320200301
52420301044103211032
635223332313231105
72214036266464133
oct240111074704061
942861574510015
1011004930132017
11356318a186088
1212989b4a02495
1361a9b695c233
142a08da24b053
151413e440e1b2
hexa0248f38831

11004930132017 has 2 divisors, whose sum is σ = 11004930132018. Its totient is φ = 11004930132016.

The previous prime is 11004930132011. The next prime is 11004930132043. The reversal of 11004930132017 is 71023103940011.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11004671694241 + 258437776 = 3317329^2 + 16076^2 .

It is a cyclic number.

It is not a de Polignac number, because 11004930132017 - 210 = 11004930130993 is a prime.

It is a super-2 number, since 2×110049301320172 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (11004930132011) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5502465066008 + 5502465066009.

It is an arithmetic number, because the mean of its divisors is an integer number (5502465066009).

Almost surely, 211004930132017 is an apocalyptic number.

It is an amenable number.

11004930132017 is a deficient number, since it is larger than the sum of its proper divisors (1).

11004930132017 is an equidigital number, since it uses as much as digits as its factorization.

11004930132017 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4536, while the sum is 32.

The spelling of 11004930132017 in words is "eleven trillion, four billion, nine hundred thirty million, one hundred thirty-two thousand, seventeen".