Base | Representation |
---|---|
bin | 10000100000011010001… |
… | …001110011100010010011 |
3 | 11000102212000122200100021 |
4 | 100200122021303202103 |
5 | 122041031344103042 |
6 | 2225032255004311 |
7 | 144644156066215 |
oct | 20403211634223 |
9 | 4012760580307 |
10 | 1134310144147 |
11 | 3a8070561124 |
12 | 163a06209697 |
13 | 82c71101c0b |
14 | 3cc8811a5b5 |
15 | 1e78cbb4867 |
hex | 1081a273893 |
1134310144147 has 2 divisors, whose sum is σ = 1134310144148. Its totient is φ = 1134310144146.
The previous prime is 1134310144133. The next prime is 1134310144213. The reversal of 1134310144147 is 7414410134311.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 1134310144147 - 215 = 1134310111379 is a prime.
It is a super-2 number, since 2×11343101441472 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1134010144147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 567155072073 + 567155072074.
It is an arithmetic number, because the mean of its divisors is an integer number (567155072074).
Almost surely, 21134310144147 is an apocalyptic number.
1134310144147 is a deficient number, since it is larger than the sum of its proper divisors (1).
1134310144147 is an equidigital number, since it uses as much as digits as its factorization.
1134310144147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 16128, while the sum is 34.
Adding to 1134310144147 its reverse (7414410134311), we get a palindrome (8548720278458).
The spelling of 1134310144147 in words is "one trillion, one hundred thirty-four billion, three hundred ten million, one hundred forty-four thousand, one hundred forty-seven".
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