Base | Representation |
---|---|
bin | 1010100000100011001101… |
… | …0100100011111000111111 |
3 | 1111220120201122021110121012 |
4 | 2220020303110203320333 |
5 | 3003301223200213033 |
6 | 40323552134225435 |
7 | 2301525616331102 |
oct | 250106324437077 |
9 | 44816648243535 |
10 | 11554323054143 |
11 | 3755185163a81 |
12 | 136737b4b527b |
13 | 65a74cb4363c |
14 | 2bd3372d4739 |
15 | 15084b46c248 |
hex | a8233523e3f |
11554323054143 has 2 divisors, whose sum is σ = 11554323054144. Its totient is φ = 11554323054142.
The previous prime is 11554323054139. The next prime is 11554323054167. The reversal of 11554323054143 is 34145032345511.
11554323054143 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 11554323054143 - 22 = 11554323054139 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 11554323054143.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11554323054193) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5777161527071 + 5777161527072.
It is an arithmetic number, because the mean of its divisors is an integer number (5777161527072).
Almost surely, 211554323054143 is an apocalyptic number.
11554323054143 is a deficient number, since it is larger than the sum of its proper divisors (1).
11554323054143 is an equidigital number, since it uses as much as digits as its factorization.
11554323054143 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 432000, while the sum is 41.
Adding to 11554323054143 its reverse (34145032345511), we get a palindrome (45699355399654).
The spelling of 11554323054143 in words is "eleven trillion, five hundred fifty-four billion, three hundred twenty-three million, fifty-four thousand, one hundred forty-three".
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