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131212214432207 is a prime number
BaseRepresentation
bin11101110101011000111001…
…011011100111000111001111
3122012120202122211022211210022
4131311120321123213013033
5114144240103343312312
61143022020440132355
736431524664315156
oct3565307133470717
9565522584284708
10131212214432207
1138898873395a97
12128719811670bb
13582a349572a6c
1424589c29a259d
1510281e6760172
hex7756396e71cf

131212214432207 has 2 divisors, whose sum is σ = 131212214432208. Its totient is φ = 131212214432206.

The previous prime is 131212214432083. The next prime is 131212214432239. The reversal of 131212214432207 is 702234412212131.

It is a strong prime.

It is an emirp because it is prime and its reverse (702234412212131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131212214432207 - 210 = 131212214431183 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (131212214432707) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (29) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65606107216103 + 65606107216104.

It is an arithmetic number, because the mean of its divisors is an integer number (65606107216104).

Almost surely, 2131212214432207 is an apocalyptic number.

131212214432207 is a deficient number, since it is larger than the sum of its proper divisors (1).

131212214432207 is an equidigital number, since it uses as much as digits as its factorization.

131212214432207 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 32256, while the sum is 35.

Adding to 131212214432207 its reverse (702234412212131), we get a palindrome (833446626644338).

The spelling of 131212214432207 in words is "one hundred thirty-one trillion, two hundred twelve billion, two hundred fourteen million, four hundred thirty-two thousand, two hundred seven".