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2143312041313 is a prime number
BaseRepresentation
bin11111001100000111010…
…110100101010101100001
321120220021002201212021001
4133030013112211111201
5240104000340310223
64320342435232001
7310564161520342
oct37140726452541
97526232655231
102143312041313
11756a78853653
122a747aa48601
13127162363411
1475a45d100c9
153ab4487e9ad
hex1f3075a5561

2143312041313 has 2 divisors, whose sum is σ = 2143312041314. Its totient is φ = 2143312041312.

The previous prime is 2143312041281. The next prime is 2143312041323. The reversal of 2143312041313 is 3131402133412.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1798152266304 + 345159775009 = 1340952^2 + 587503^2 .

It is a cyclic number.

It is not a de Polignac number, because 2143312041313 - 25 = 2143312041281 is a prime.

It is not a weakly prime, because it can be changed into another prime (2143312041323) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1071656020656 + 1071656020657.

It is an arithmetic number, because the mean of its divisors is an integer number (1071656020657).

Almost surely, 22143312041313 is an apocalyptic number.

It is an amenable number.

2143312041313 is a deficient number, since it is larger than the sum of its proper divisors (1).

2143312041313 is an equidigital number, since it uses as much as digits as its factorization.

2143312041313 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 5184, while the sum is 28.

Adding to 2143312041313 its reverse (3131402133412), we get a palindrome (5274714174725).

The spelling of 2143312041313 in words is "two trillion, one hundred forty-three billion, three hundred twelve million, forty-one thousand, three hundred thirteen".