Base | Representation |
---|---|
bin | 10100100010011000101… |
… | …101110111100110100011 |
3 | 11222220211201110012001022 |
4 | 110202120231313212203 |
5 | 141110331211343042 |
6 | 3000203022000055 |
7 | 203651420154116 |
oct | 24423055674643 |
9 | 4886751405038 |
10 | 1411311434147 |
11 | 4a459656a17a |
12 | 1a963141602b |
13 | a3117203884 |
14 | 4c4449c857d |
15 | 26aa121acd2 |
hex | 14898b779a3 |
1411311434147 has 2 divisors, whose sum is σ = 1411311434148. Its totient is φ = 1411311434146.
The previous prime is 1411311434119. The next prime is 1411311434153. The reversal of 1411311434147 is 7414341131141.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1411311434147 - 218 = 1411311172003 is a prime.
It is not a weakly prime, because it can be changed into another prime (1411311434107) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 705655717073 + 705655717074.
It is an arithmetic number, because the mean of its divisors is an integer number (705655717074).
Almost surely, 21411311434147 is an apocalyptic number.
1411311434147 is a deficient number, since it is larger than the sum of its proper divisors (1).
1411311434147 is an equidigital number, since it uses as much as digits as its factorization.
1411311434147 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 16128, while the sum is 35.
Adding to 1411311434147 its reverse (7414341131141), we get a palindrome (8825652565288).
The spelling of 1411311434147 in words is "one trillion, four hundred eleven billion, three hundred eleven million, four hundred thirty-four thousand, one hundred forty-seven".
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