Base | Representation |
---|---|
bin | 11110000000011000110… |
… | …10000111100000100011 |
3 | 10122120012012000001121001 |
4 | 33000030122013200203 |
5 | 113342442040333011 |
6 | 2105345055012431 |
7 | 134326102105531 |
oct | 17001432074043 |
9 | 3576165001531 |
10 | 1031000324131 |
11 | 368276894288 |
12 | 147993b0a117 |
13 | 762b89824bb |
14 | 37c87685151 |
15 | 1bc430d77c1 |
hex | f00c687823 |
1031000324131 has 2 divisors, whose sum is σ = 1031000324132. Its totient is φ = 1031000324130.
The previous prime is 1031000324111. The next prime is 1031000324153. The reversal of 1031000324131 is 1314230001301.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1031000324131 - 211 = 1031000322083 is a prime.
It is a super-2 number, since 2×10310003241312 (a number of 25 digits) contains 22 as substring.
It is a junction number, because it is equal to n+sod(n) for n = 1031000324099 and 1031000324108.
It is not a weakly prime, because it can be changed into another prime (1031000324111) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 515500162065 + 515500162066.
It is an arithmetic number, because the mean of its divisors is an integer number (515500162066).
Almost surely, 21031000324131 is an apocalyptic number.
1031000324131 is a deficient number, since it is larger than the sum of its proper divisors (1).
1031000324131 is an equidigital number, since it uses as much as digits as its factorization.
1031000324131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 216, while the sum is 19.
Adding to 1031000324131 its reverse (1314230001301), we get a palindrome (2345230325432).
The spelling of 1031000324131 in words is "one trillion, thirty-one billion, three hundred twenty-four thousand, one hundred thirty-one".
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