Base | Representation |
---|---|
bin | 111100000111011001100… |
… | …110100101101010001011 |
3 | 112121221010110110102211211 |
4 | 330013121212211222023 |
5 | 1020141010232434201 |
6 | 12441450245233551 |
7 | 604314602352043 |
oct | 74073146455213 |
9 | 15557113412754 |
10 | 4131114343051 |
11 | 1352aa5392173 |
12 | 568779a028b7 |
13 | 23c740803547 |
14 | 103d387dc723 |
15 | 726d6a89651 |
hex | 3c1d99a5a8b |
4131114343051 has 2 divisors, whose sum is σ = 4131114343052. Its totient is φ = 4131114343050.
The previous prime is 4131114343021. The next prime is 4131114343063. The reversal of 4131114343051 is 1503434111314.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 4131114343051 - 29 = 4131114342539 is a prime.
It is a super-2 number, since 2×41311143430512 (a number of 26 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (4131114343021) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2065557171525 + 2065557171526.
It is an arithmetic number, because the mean of its divisors is an integer number (2065557171526).
Almost surely, 24131114343051 is an apocalyptic number.
4131114343051 is a deficient number, since it is larger than the sum of its proper divisors (1).
4131114343051 is an equidigital number, since it uses as much as digits as its factorization.
4131114343051 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8640, while the sum is 31.
Adding to 4131114343051 its reverse (1503434111314), we get a palindrome (5634548454365).
The spelling of 4131114343051 in words is "four trillion, one hundred thirty-one billion, one hundred fourteen million, three hundred forty-three thousand, fifty-one".
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