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213442131002117 is a prime number
BaseRepresentation
bin110000100001111111011110…
…011001100000001100000101
31000222201212110002121121200112
4300201333132121200030011
5210434013441014031432
62033541543351522405
762646453660516203
oct6041773631401405
91028655402547615
10213442131002117
1162011332586956
121bb32621989405
139213677c8bb84
143a9c923416073
1519a21bdc447b2
hexc21fde660305

213442131002117 has 2 divisors, whose sum is σ = 213442131002118. Its totient is φ = 213442131002116.

The previous prime is 213442131002081. The next prime is 213442131002119. The reversal of 213442131002117 is 711200131244312.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 185884156316356 + 27557974685761 = 13633934^2 + 5249569^2 .

It is a cyclic number.

It is not a de Polignac number, because 213442131002117 - 218 = 213442130739973 is a prime.

Together with 213442131002119, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (213442131002119) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 106721065501058 + 106721065501059.

It is an arithmetic number, because the mean of its divisors is an integer number (106721065501059).

Almost surely, 2213442131002117 is an apocalyptic number.

It is an amenable number.

213442131002117 is a deficient number, since it is larger than the sum of its proper divisors (1).

213442131002117 is an equidigital number, since it uses as much as digits as its factorization.

213442131002117 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 8064, while the sum is 32.

Adding to 213442131002117 its reverse (711200131244312), we get a palindrome (924642262246429).

The spelling of 213442131002117 in words is "two hundred thirteen trillion, four hundred forty-two billion, one hundred thirty-one million, two thousand, one hundred seventeen".