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113130031309 is a prime number
BaseRepresentation
bin110100101011100010…
…0110100100011001101
3101211000020010001010121
41221113010310203031
53323142242000214
6123545441104541
711113316254111
oct1512704644315
9354006101117
10113130031309
1143a83a53a49
1219b13028151
13a88bb0b239
145692b97741
152e21cc4424
hex1a571348cd

113130031309 has 2 divisors, whose sum is σ = 113130031310. Its totient is φ = 113130031308.

The previous prime is 113130031307. The next prime is 113130031331. The reversal of 113130031309 is 903130031311.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 110837058084 + 2292973225 = 332922^2 + 47885^2 .

It is an emirp because it is prime and its reverse (903130031311) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 113130031309 - 21 = 113130031307 is a prime.

Together with 113130031307, it forms a pair of twin primes.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (113130031303) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56565015654 + 56565015655.

It is an arithmetic number, because the mean of its divisors is an integer number (56565015655).

Almost surely, 2113130031309 is an apocalyptic number.

It is an amenable number.

113130031309 is a deficient number, since it is larger than the sum of its proper divisors (1).

113130031309 is an equidigital number, since it uses as much as digits as its factorization.

113130031309 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 729, while the sum is 25.

The spelling of 113130031309 in words is "one hundred thirteen billion, one hundred thirty million, thirty-one thousand, three hundred nine".