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11234200313153 is a prime number
BaseRepresentation
bin1010001101111010101010…
…0001001101110101000001
31110202222102201212021121202
42203132222201031311001
52433030120240010103
635520530445030545
72236433646043166
oct243365241156501
943688381767552
1011234200313153
113641441224624
12131531ab55455
136364c40936cc
142aba4987a06d
15147362399988
hexa37aa84dd41

11234200313153 has 2 divisors, whose sum is σ = 11234200313154. Its totient is φ = 11234200313152.

The previous prime is 11234200313131. The next prime is 11234200313207. The reversal of 11234200313153 is 35131300243211.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 7296087070129 + 3938113243024 = 2701127^2 + 1984468^2 .

It is a cyclic number.

It is not a de Polignac number, because 11234200313153 - 28 = 11234200312897 is a prime.

It is not a weakly prime, because it can be changed into another prime (11234200318153) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5617100156576 + 5617100156577.

It is an arithmetic number, because the mean of its divisors is an integer number (5617100156577).

Almost surely, 211234200313153 is an apocalyptic number.

It is an amenable number.

11234200313153 is a deficient number, since it is larger than the sum of its proper divisors (1).

11234200313153 is an equidigital number, since it uses as much as digits as its factorization.

11234200313153 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 6480, while the sum is 29.

Adding to 11234200313153 its reverse (35131300243211), we get a palindrome (46365500556364).

The spelling of 11234200313153 in words is "eleven trillion, two hundred thirty-four billion, two hundred million, three hundred thirteen thousand, one hundred fifty-three".