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333142343241317 is a prime number
BaseRepresentation
bin100101110111111011011111…
…1000001110100111001100101
31121200120002121200211111022222
41023233312333001310321211
5322131201004332210232
63140311251340402125
7130112511554052131
oct11357667701647145
91550502550744288
10333142343241317
1197170968836265
1231445274846345
13113b725a53b06c
145c3a2655d03c1
15287abe4661d12
hex12efdbf074e65

333142343241317 has 2 divisors, whose sum is σ = 333142343241318. Its totient is φ = 333142343241316.

The previous prime is 333142343241299. The next prime is 333142343241377. The reversal of 333142343241317 is 713142343241333.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 193634850495076 + 139507492746241 = 13915274^2 + 11811329^2 .

It is a cyclic number.

It is not a de Polignac number, because 333142343241317 - 210 = 333142343240293 is a prime.

It is a super-2 number, since 2×3331423432413172 (a number of 30 digits) contains 22 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (333142343241377) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 166571171620658 + 166571171620659.

It is an arithmetic number, because the mean of its divisors is an integer number (166571171620659).

Almost surely, 2333142343241317 is an apocalyptic number.

It is an amenable number.

333142343241317 is a deficient number, since it is larger than the sum of its proper divisors (1).

333142343241317 is an equidigital number, since it uses as much as digits as its factorization.

333142343241317 is an evil number, because the sum of its binary digits is even.

The product of its digits is 1306368, while the sum is 44.

The spelling of 333142343241317 in words is "three hundred thirty-three trillion, one hundred forty-two billion, three hundred forty-three million, two hundred forty-one thousand, three hundred seventeen".