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12836348327701 is a prime number
BaseRepresentation
bin1011101011001011000111…
…1110110111011100010101
31200110010212012000012121101
42322302301332313130111
53140302320132441301
643144534221150101
72463252451342303
oct272626176673425
950403765005541
1012836348327701
1140a9956a96148
12153392b0bb331
13721602019072
143253d5551273
15173d8232c701
hexbacb1fb7715

12836348327701 has 2 divisors, whose sum is σ = 12836348327702. Its totient is φ = 12836348327700.

The previous prime is 12836348327683. The next prime is 12836348327711. The reversal of 12836348327701 is 10772384363821.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 7051409391601 + 5784938936100 = 2655449^2 + 2405190^2 .

It is a cyclic number.

It is not a de Polignac number, because 12836348327701 - 29 = 12836348327189 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (12836348327711) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6418174163850 + 6418174163851.

It is an arithmetic number, because the mean of its divisors is an integer number (6418174163851).

Almost surely, 212836348327701 is an apocalyptic number.

It is an amenable number.

12836348327701 is a deficient number, since it is larger than the sum of its proper divisors (1).

12836348327701 is an equidigital number, since it uses as much as digits as its factorization.

12836348327701 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 8128512, while the sum is 55.

The spelling of 12836348327701 in words is "twelve trillion, eight hundred thirty-six billion, three hundred forty-eight million, three hundred twenty-seven thousand, seven hundred one".