Base | Representation |
---|---|
bin | 101110000011001011100… |
… | …011100111011001011001 |
3 | 102012112011012211001111211 |
4 | 232003023203213121121 |
5 | 403321404311303423 |
6 | 10421431210023121 |
7 | 444425351014462 |
oct | 56031343473131 |
9 | 12175135731454 |
10 | 3164511041113 |
11 | 1010074a90894 |
12 | 431378417aa1 |
13 | 19c54853aa06 |
14 | ad23d98bd69 |
15 | 574b225110d |
hex | 2e0cb8e7659 |
3164511041113 has 2 divisors, whose sum is σ = 3164511041114. Its totient is φ = 3164511041112.
The previous prime is 3164511041077. The next prime is 3164511041143. The reversal of 3164511041113 is 3111401154613.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 2954538141129 + 209972899984 = 1718877^2 + 458228^2 .
It is a cyclic number.
It is not a de Polignac number, because 3164511041113 - 29 = 3164511040601 is a prime.
It is not a weakly prime, because it can be changed into another prime (3164511041143) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1582255520556 + 1582255520557.
It is an arithmetic number, because the mean of its divisors is an integer number (1582255520557).
Almost surely, 23164511041113 is an apocalyptic number.
It is an amenable number.
3164511041113 is a deficient number, since it is larger than the sum of its proper divisors (1).
3164511041113 is an equidigital number, since it uses as much as digits as its factorization.
3164511041113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4320, while the sum is 31.
Adding to 3164511041113 its reverse (3111401154613), we get a palindrome (6275912195726).
The spelling of 3164511041113 in words is "three trillion, one hundred sixty-four billion, five hundred eleven million, forty-one thousand, one hundred thirteen".
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