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131121304513471 is a prime number
BaseRepresentation
bin11101110100000100001110…
…110001110010101110111111
3122012021002000011021110211001
4131310010032301302232333
5114141242412423412341
61142512143513111131
736422126105622256
oct3564041661625677
9565232004243731
10131121304513471
1138863262088548
121285822b6b84a7
1358218bc09353c
14245443acc579d
151025b75597d31
hex77410ec72bbf

131121304513471 has 2 divisors, whose sum is σ = 131121304513472. Its totient is φ = 131121304513470.

The previous prime is 131121304513351. The next prime is 131121304513499. The reversal of 131121304513471 is 174315403121131.

It is a strong prime.

It is an emirp because it is prime and its reverse (174315403121131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 131121304513471 - 219 = 131121303989183 is a prime.

It is a super-3 number, since 3×1311213045134713 (a number of 43 digits) contains 333 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 131121304513471.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (131121304513271) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65560652256735 + 65560652256736.

It is an arithmetic number, because the mean of its divisors is an integer number (65560652256736).

Almost surely, 2131121304513471 is an apocalyptic number.

131121304513471 is a deficient number, since it is larger than the sum of its proper divisors (1).

131121304513471 is an equidigital number, since it uses as much as digits as its factorization.

131121304513471 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 30240, while the sum is 37.

The spelling of 131121304513471 in words is "one hundred thirty-one trillion, one hundred twenty-one billion, three hundred four million, five hundred thirteen thousand, four hundred seventy-one".