Base | Representation |
---|---|
bin | 100101111110111111001110… |
… | …0000111011111111000000011 |
3 | 1121210222211120000122120220222 |
4 | 1023331332130013133320003 |
5 | 322243044214023300103 |
6 | 3142333043223053255 |
7 | 130242553144435331 |
oct | 11375763407377003 |
9 | 1553884500576828 |
10 | 334112420134403 |
11 | 97505310967029 |
12 | 315812855a122b |
13 | 11457887354981 |
14 | 5c711ad8c6b51 |
15 | 289606dddc638 |
hex | 12fdf9c1dfe03 |
334112420134403 has 2 divisors, whose sum is σ = 334112420134404. Its totient is φ = 334112420134402.
The previous prime is 334112420134397. The next prime is 334112420134469. The reversal of 334112420134403 is 304431024211433.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 334112420134403 - 222 = 334112415940099 is a prime.
It is a super-2 number, since 2×3341124201344032 (a number of 30 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (334112420139403) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 167056210067201 + 167056210067202.
It is an arithmetic number, because the mean of its divisors is an integer number (167056210067202).
Almost surely, 2334112420134403 is an apocalyptic number.
334112420134403 is a deficient number, since it is larger than the sum of its proper divisors (1).
334112420134403 is an equidigital number, since it uses as much as digits as its factorization.
334112420134403 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 82944, while the sum is 35.
Adding to 334112420134403 its reverse (304431024211433), we get a palindrome (638543444345836).
The spelling of 334112420134403 in words is "three hundred thirty-four trillion, one hundred twelve billion, four hundred twenty million, one hundred thirty-four thousand, four hundred three".
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