Base | Representation |
---|---|
bin | 100110101000101010101… |
… | …0101100110110100011011 |
3 | 200210122001211002011011101 |
4 | 1031101111111212310123 |
5 | 1143444400343231433 |
6 | 15143215134412231 |
7 | 1055431013243401 |
oct | 115212525466433 |
9 | 20718054064141 |
10 | 5310011305243 |
11 | 1767a65a4a647 |
12 | 719148061077 |
13 | 2c696903b018 |
14 | 145012351471 |
15 | 931d3c7307d |
hex | 4d455566d1b |
5310011305243 has 4 divisors (see below), whose sum is σ = 5316038560728. Its totient is φ = 5303984049760.
The previous prime is 5310011305217. The next prime is 5310011305289. The reversal of 5310011305243 is 3425031100135.
It is a happy number.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-5310011305243 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (5310011302243) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3013626421 + ... + 3013628182.
It is an arithmetic number, because the mean of its divisors is an integer number (1329009640182).
Almost surely, 25310011305243 is an apocalyptic number.
5310011305243 is a deficient number, since it is larger than the sum of its proper divisors (6027255485).
5310011305243 is an equidigital number, since it uses as much as digits as its factorization.
5310011305243 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 6027255484.
The product of its (nonzero) digits is 5400, while the sum is 28.
Adding to 5310011305243 its reverse (3425031100135), we get a palindrome (8735042405378).
The spelling of 5310011305243 in words is "five trillion, three hundred ten billion, eleven million, three hundred five thousand, two hundred forty-three".
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