Base | Representation |
---|---|
bin | 101110000100010011011… |
… | …1001010001001111011101 |
3 | 211102021112112201120102211 |
4 | 1130020212321101033131 |
5 | 1312213234142410331 |
6 | 21244342031120421 |
7 | 1222300562634544 |
oct | 134104671211735 |
9 | 24367475646384 |
10 | 6331434341341 |
11 | 202116769764a |
12 | 8630a8673111 |
13 | 36c089767285 |
14 | 17c62ba6b15b |
15 | aea6614a8b1 |
hex | 5c226e513dd |
6331434341341 has 2 divisors, whose sum is σ = 6331434341342. Its totient is φ = 6331434341340.
The previous prime is 6331434341311. The next prime is 6331434341357. The reversal of 6331434341341 is 1431434341336.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 3858533419225 + 2472900922116 = 1964315^2 + 1572546^2 .
It is a cyclic number.
It is not a de Polignac number, because 6331434341341 - 213 = 6331434333149 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 6331434341294 and 6331434341303.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (6331434341311) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3165717170670 + 3165717170671.
It is an arithmetic number, because the mean of its divisors is an integer number (3165717170671).
Almost surely, 26331434341341 is an apocalyptic number.
It is an amenable number.
6331434341341 is a deficient number, since it is larger than the sum of its proper divisors (1).
6331434341341 is an equidigital number, since it uses as much as digits as its factorization.
6331434341341 is an evil number, because the sum of its binary digits is even.
The product of its digits is 373248, while the sum is 40.
Adding to 6331434341341 its reverse (1431434341336), we get a palindrome (7762868682677).
The spelling of 6331434341341 in words is "six trillion, three hundred thirty-one billion, four hundred thirty-four million, three hundred forty-one thousand, three hundred forty-one".
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