Base | Representation |
---|---|
bin | 101011110100011010110… |
… | …001010100101111010001 |
3 | 101122212111011221220122202 |
4 | 223310122301110233101 |
5 | 343313440112241423 |
6 | 10223200321012545 |
7 | 430360603662152 |
oct | 53643261245721 |
9 | 11585434856582 |
10 | 3011221212113 |
11 | a610629654a4 |
12 | 407717a84155 |
13 | 18ac5952a89a |
14 | a5a5b30b929 |
15 | 534de92c728 |
hex | 2bd1ac54bd1 |
3011221212113 has 2 divisors, whose sum is σ = 3011221212114. Its totient is φ = 3011221212112.
The previous prime is 3011221212091. The next prime is 3011221212137. The reversal of 3011221212113 is 3112121221103.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2981106121744 + 30115090369 = 1726588^2 + 173537^2 .
It is a cyclic number.
It is not a de Polignac number, because 3011221212113 - 218 = 3011220949969 is a prime.
It is a Sophie Germain prime.
It is a Curzon number.
It is not a weakly prime, because it can be changed into another prime (3011221212143) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1505610606056 + 1505610606057.
It is an arithmetic number, because the mean of its divisors is an integer number (1505610606057).
Almost surely, 23011221212113 is an apocalyptic number.
It is an amenable number.
3011221212113 is a deficient number, since it is larger than the sum of its proper divisors (1).
3011221212113 is an equidigital number, since it uses as much as digits as its factorization.
3011221212113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 144, while the sum is 20.
Adding to 3011221212113 its reverse (3112121221103), we get a palindrome (6123342433216).
The spelling of 3011221212113 in words is "three trillion, eleven billion, two hundred twenty-one million, two hundred twelve thousand, one hundred thirteen".
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