Base | Representation |
---|---|
bin | 11000101101100110111011… |
… | …00111101110000110000011 |
3 | 21010102012110000102112201001 |
4 | 30112303131213232012003 |
5 | 24110331240431444042 |
6 | 311325043043352431 |
7 | 14306125111611325 |
oct | 1426633547560603 |
9 | 233365400375631 |
10 | 54343644406147 |
11 | 163520125932a8 |
12 | 61181b4682717 |
13 | 2442778459bb5 |
14 | d5c364285615 |
15 | 643909be4bb7 |
hex | 316cdd9ee183 |
54343644406147 has 8 divisors (see below), whose sum is σ = 55154977866512. Its totient is φ = 53532317836800.
The previous prime is 54343644406111. The next prime is 54343644406151. The reversal of 54343644406147 is 74160444634345.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-54343644406147 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (54343644206147) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 15433147 + ... + 18624427.
It is an arithmetic number, because the mean of its divisors is an integer number (6894372233314).
Almost surely, 254343644406147 is an apocalyptic number.
54343644406147 is a deficient number, since it is larger than the sum of its proper divisors (811333460365).
54343644406147 is a wasteful number, since it uses less digits than its factorization.
54343644406147 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3445509.
The product of its (nonzero) digits is 46448640, while the sum is 55.
The spelling of 54343644406147 in words is "fifty-four trillion, three hundred forty-three billion, six hundred forty-four million, four hundred six thousand, one hundred forty-seven".
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