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353151044117 is a prime number
BaseRepresentation
bin1010010001110010111…
…00010100101000010101
31020202112200222211021122
411020321130110220111
521241223131402432
6430122500253325
734341261226544
oct5107134245025
91222480884248
10353151044117
11126852615443
12585396a1245
13273c0659081
141314215615b
1592bd9d1612
hex5239714a15

353151044117 has 2 divisors, whose sum is σ = 353151044118. Its totient is φ = 353151044116.

The previous prime is 353151044101. The next prime is 353151044123. The reversal of 353151044117 is 711440151353.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 315528234961 + 37622809156 = 561719^2 + 193966^2 .

It is an emirp because it is prime and its reverse (711440151353) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 353151044117 - 24 = 353151044101 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (353151044167) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 176575522058 + 176575522059.

It is an arithmetic number, because the mean of its divisors is an integer number (176575522059).

Almost surely, 2353151044117 is an apocalyptic number.

It is an amenable number.

353151044117 is a deficient number, since it is larger than the sum of its proper divisors (1).

353151044117 is an equidigital number, since it uses as much as digits as its factorization.

353151044117 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 25200, while the sum is 35.

The spelling of 353151044117 in words is "three hundred fifty-three billion, one hundred fifty-one million, forty-four thousand, one hundred seventeen".