Base | Representation |
---|---|
bin | 110101001110110010110100… |
… | …111100011101110110010011 |
3 | 1010200220222212101022101100221 |
4 | 311032302310330131312103 |
5 | 221141202123424321042 |
6 | 2145530025053253511 |
7 | 100212054121461502 |
oct | 6516626474356623 |
9 | 1120828771271327 |
10 | 234113113120147 |
11 | 686608a5049348 |
12 | 22310833a94297 |
13 | a082a05963581 |
14 | 41b51c6b44839 |
15 | 1c0ec453a6d67 |
hex | d4ecb4f1dd93 |
234113113120147 has 2 divisors, whose sum is σ = 234113113120148. Its totient is φ = 234113113120146.
The previous prime is 234113113120109. The next prime is 234113113120163. The reversal of 234113113120147 is 741021311311432.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 234113113120147 - 211 = 234113113118099 is a prime.
It is not a weakly prime, because it can be changed into another prime (234113113120247) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 117056556560073 + 117056556560074.
It is an arithmetic number, because the mean of its divisors is an integer number (117056556560074).
Almost surely, 2234113113120147 is an apocalyptic number.
234113113120147 is a deficient number, since it is larger than the sum of its proper divisors (1).
234113113120147 is an equidigital number, since it uses as much as digits as its factorization.
234113113120147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 12096, while the sum is 34.
Adding to 234113113120147 its reverse (741021311311432), we get a palindrome (975134424431579).
The spelling of 234113113120147 in words is "two hundred thirty-four trillion, one hundred thirteen billion, one hundred thirteen million, one hundred twenty thousand, one hundred forty-seven".
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