Base | Representation |
---|---|
bin | 10010111110101000011… |
… | …100011011010110100011 |
3 | 11121200101000111012122121 |
4 | 102332220130123112203 |
5 | 132332000441140133 |
6 | 2435050340045111 |
7 | 163140214365241 |
oct | 22765034332643 |
9 | 4550330435577 |
10 | 1304201115043 |
11 | 4631215900a2 |
12 | 19091a393797 |
13 | 95ca75a3828 |
14 | 471a34d1791 |
15 | 23dd2b7502d |
hex | 12fa871b5a3 |
1304201115043 has 2 divisors, whose sum is σ = 1304201115044. Its totient is φ = 1304201115042.
The previous prime is 1304201115023. The next prime is 1304201115053. The reversal of 1304201115043 is 3405111024031.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1304201115043 - 217 = 1304200983971 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 1304201114999 and 1304201115017.
It is not a weakly prime, because it can be changed into another prime (1304201115023) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 652100557521 + 652100557522.
It is an arithmetic number, because the mean of its divisors is an integer number (652100557522).
Almost surely, 21304201115043 is an apocalyptic number.
1304201115043 is a deficient number, since it is larger than the sum of its proper divisors (1).
1304201115043 is an equidigital number, since it uses as much as digits as its factorization.
1304201115043 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1440, while the sum is 25.
Adding to 1304201115043 its reverse (3405111024031), we get a palindrome (4709312139074).
The spelling of 1304201115043 in words is "one trillion, three hundred four billion, two hundred one million, one hundred fifteen thousand, forty-three".
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