Search a number
-
+
94331211212857 is a prime number
BaseRepresentation
bin10101011100101100110010…
…010101011110100000111001
3110100222222100022002220012011
4111130230302111132200321
544331010310032302412
6532343054333245521
725604130231401203
oct2534546225364071
9410888308086164
1094331211212857
112806970757a213
12a6b602b9338a1
134083519bc8c48
1419419253bb773
15ad8b881227a7
hex55cb3255e839

94331211212857 has 2 divisors, whose sum is σ = 94331211212858. Its totient is φ = 94331211212856.

The previous prime is 94331211212837. The next prime is 94331211212873. The reversal of 94331211212857 is 75821211213349.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 93754369443856 + 576841769001 = 9682684^2 + 759501^2 .

It is a cyclic number.

It is not a de Polignac number, because 94331211212857 - 211 = 94331211210809 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 94331211212857.

It is not a weakly prime, because it can be changed into another prime (94331211212837) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 47165605606428 + 47165605606429.

It is an arithmetic number, because the mean of its divisors is an integer number (47165605606429).

Almost surely, 294331211212857 is an apocalyptic number.

It is an amenable number.

94331211212857 is a deficient number, since it is larger than the sum of its proper divisors (1).

94331211212857 is an equidigital number, since it uses as much as digits as its factorization.

94331211212857 is an evil number, because the sum of its binary digits is even.

The product of its digits is 725760, while the sum is 49.

The spelling of 94331211212857 in words is "ninety-four trillion, three hundred thirty-one billion, two hundred eleven million, two hundred twelve thousand, eight hundred fifty-seven".