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135921993017 is a prime number
BaseRepresentation
bin111111010010110010…
…1010000010100111001
3110222211122012122211112
41332211211100110321
54211332012234032
6142235224010105
712551162610056
oct1764545202471
9428748178745
10135921993017
115270a470138
1222414000935
13ca81a5a405
146815bb002d
153807bdd1b2
hex1fa5950539

135921993017 has 2 divisors, whose sum is σ = 135921993018. Its totient is φ = 135921993016.

The previous prime is 135921992903. The next prime is 135921993019. The reversal of 135921993017 is 710399129531.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 108335772736 + 27586220281 = 329144^2 + 166091^2 .

It is a cyclic number.

It is not a de Polignac number, because 135921993017 - 28 = 135921992761 is a prime.

It is a super-3 number, since 3×1359219930173 (a number of 34 digits) contains 333 as substring.

Together with 135921993019, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (135921993019) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67960996508 + 67960996509.

It is an arithmetic number, because the mean of its divisors is an integer number (67960996509).

Almost surely, 2135921993017 is an apocalyptic number.

It is an amenable number.

135921993017 is a deficient number, since it is larger than the sum of its proper divisors (1).

135921993017 is an equidigital number, since it uses as much as digits as its factorization.

135921993017 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 459270, while the sum is 50.

The spelling of 135921993017 in words is "one hundred thirty-five billion, nine hundred twenty-one million, nine hundred ninety-three thousand, seventeen".