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94331211213077 is a prime number
BaseRepresentation
bin10101011100101100110010…
…010101011110100100010101
3110100222222100022002220111022
4111130230302111132210111
544331010310032304302
6532343054333250525
725604130231401636
oct2534546225364425
9410888308086438
1094331211213077
112806970757a3a3
12a6b602b933a45
134083519bc9087
1419419253bb88d
15ad8b881228a2
hex55cb3255e915

94331211213077 has 2 divisors, whose sum is σ = 94331211213078. Its totient is φ = 94331211213076.

The previous prime is 94331211213061. The next prime is 94331211213079. The reversal of 94331211213077 is 77031211213349.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 91983617274481 + 2347593938596 = 9590809^2 + 1532186^2 .

It is a cyclic number.

It is not a de Polignac number, because 94331211213077 - 24 = 94331211213061 is a prime.

It is a super-2 number, since 2×943312112130772 (a number of 29 digits) contains 22 as substring.

Together with 94331211213079, 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 (94331211213079) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 47165605606538 + 47165605606539.

It is an arithmetic number, because the mean of its divisors is an integer number (47165605606539).

Almost surely, 294331211213077 is an apocalyptic number.

It is an amenable number.

94331211213077 is a deficient number, since it is larger than the sum of its proper divisors (1).

94331211213077 is an equidigital number, since it uses as much as digits as its factorization.

94331211213077 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 190512, while the sum is 44.

The spelling of 94331211213077 in words is "ninety-four trillion, three hundred thirty-one billion, two hundred eleven million, two hundred thirteen thousand, seventy-seven".