Base | Representation |
---|---|
bin | 101101010000100111100… |
… | …100001110101011100001 |
3 | 102000100000120220001211201 |
4 | 231100213210032223201 |
5 | 401424212323143131 |
6 | 10340452043241201 |
7 | 440464103625022 |
oct | 55204744165341 |
9 | 12010016801751 |
10 | 3110220131041 |
11 | a9a0451a9571 |
12 | 422946519201 |
13 | 1973a581205a |
14 | aa76d41cc49 |
15 | 55d85d20761 |
hex | 2d42790eae1 |
3110220131041 has 2 divisors, whose sum is σ = 3110220131042. Its totient is φ = 3110220131040.
The previous prime is 3110220131027. The next prime is 3110220131107. The reversal of 3110220131041 is 1401310220113.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 2831156220816 + 279063910225 = 1682604^2 + 528265^2 .
It is an emirp because it is prime and its reverse (1401310220113) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 3110220131041 - 215 = 3110220098273 is a prime.
It is not a weakly prime, because it can be changed into another prime (3110220131021) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1555110065520 + 1555110065521.
It is an arithmetic number, because the mean of its divisors is an integer number (1555110065521).
Almost surely, 23110220131041 is an apocalyptic number.
It is an amenable number.
3110220131041 is a deficient number, since it is larger than the sum of its proper divisors (1).
3110220131041 is an equidigital number, since it uses as much as digits as its factorization.
3110220131041 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 19.
Adding to 3110220131041 its reverse (1401310220113), we get a palindrome (4511530351154).
The spelling of 3110220131041 in words is "three trillion, one hundred ten billion, two hundred twenty million, one hundred thirty-one thousand, forty-one".
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