Base | Representation |
---|---|
bin | 1111101110011010011… |
… | …00000011010001110001 |
3 | 1220122122100222122022110 |
4 | 13313031030003101301 |
5 | 32323030212303213 |
6 | 1052114435155533 |
7 | 54015320061132 |
oct | 7671514032161 |
9 | 1818570878273 |
10 | 540313400433 |
11 | 199166912964 |
12 | 88871a45ba9 |
13 | 3bc4a110681 |
14 | 1c219291c89 |
15 | e0c4dc45c3 |
hex | 7dcd303471 |
540313400433 has 16 divisors (see below), whose sum is σ = 734219531136. Its totient is φ = 353311969920.
The previous prime is 540313400423. The next prime is 540313400441. The reversal of 540313400433 is 334004313045.
It is not a de Polignac number, because 540313400433 - 218 = 540313138289 is a prime.
It is a Curzon number.
It is a self number, because there is not a number n which added to its sum of digits gives 540313400433.
It is not an unprimeable number, because it can be changed into a prime (540313400423) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 79053 + ... + 1042533.
It is an arithmetic number, because the mean of its divisors is an integer number (45888720696).
Almost surely, 2540313400433 is an apocalyptic number.
540313400433 is a gapful number since it is divisible by the number (53) formed by its first and last digit.
It is an amenable number.
540313400433 is a deficient number, since it is larger than the sum of its proper divisors (193906130703).
540313400433 is a wasteful number, since it uses less digits than its factorization.
540313400433 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 967064.
The product of its (nonzero) digits is 25920, while the sum is 30.
Adding to 540313400433 its reverse (334004313045), we get a palindrome (874317713478).
The spelling of 540313400433 in words is "five hundred forty billion, three hundred thirteen million, four hundred thousand, four hundred thirty-three".
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