Base | Representation |
---|---|
bin | 1001000110000101001… |
… | …10001011100101000001 |
3 | 1002212121211110111010102 |
4 | 10203002212023211001 |
5 | 20110001110342423 |
6 | 355321151412145 |
7 | 31402046601512 |
oct | 4430246134501 |
9 | 1085554414112 |
10 | 312502434113 |
11 | 110593528081 |
12 | 50694524055 |
13 | 236130bb40c |
14 | 111a778aa09 |
15 | 81e00c8c28 |
hex | 48c298b941 |
312502434113 has 2 divisors, whose sum is σ = 312502434114. Its totient is φ = 312502434112.
The previous prime is 312502434109. The next prime is 312502434151. The reversal of 312502434113 is 311434205213.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 245314165264 + 67188268849 = 495292^2 + 259207^2 .
It is an emirp because it is prime and its reverse (311434205213) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 312502434113 - 22 = 312502434109 is a prime.
It is not a weakly prime, because it can be changed into another prime (312502434613) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 156251217056 + 156251217057.
It is an arithmetic number, because the mean of its divisors is an integer number (156251217057).
Almost surely, 2312502434113 is an apocalyptic number.
It is an amenable number.
312502434113 is a deficient number, since it is larger than the sum of its proper divisors (1).
312502434113 is an equidigital number, since it uses as much as digits as its factorization.
312502434113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 8640, while the sum is 29.
Adding to 312502434113 its reverse (311434205213), we get a palindrome (623936639326).
The spelling of 312502434113 in words is "three hundred twelve billion, five hundred two million, four hundred thirty-four thousand, one hundred thirteen".
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