Base | Representation |
---|---|
bin | 11110000000110001110… |
… | …11001010010110101011 |
3 | 10122120201211010121100102 |
4 | 33000120323022112223 |
5 | 113343404320010011 |
6 | 2105421555500015 |
7 | 134334232015415 |
oct | 17003073122653 |
9 | 3576654117312 |
10 | 1031210313131 |
11 | 36837437a109 |
12 | 147a322b760b |
13 | 7632032634b |
14 | 37ca75079b5 |
15 | 1bc5676673b |
hex | f018eca5ab |
1031210313131 has 2 divisors, whose sum is σ = 1031210313132. Its totient is φ = 1031210313130.
The previous prime is 1031210313121. The next prime is 1031210313163. The reversal of 1031210313131 is 1313130121301.
1031210313131 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1031210313131 - 214 = 1031210296747 is a prime.
It is a super-2 number, since 2×10312103131312 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1031210313121) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 515605156565 + 515605156566.
It is an arithmetic number, because the mean of its divisors is an integer number (515605156566).
Almost surely, 21031210313131 is an apocalyptic number.
1031210313131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1031210313131 is an equidigital number, since it uses as much as digits as its factorization.
1031210313131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 162, while the sum is 20.
Adding to 1031210313131 its reverse (1313130121301), we get a palindrome (2344340434432).
The spelling of 1031210313131 in words is "one trillion, thirty-one billion, two hundred ten million, three hundred thirteen thousand, one hundred thirty-one".
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