Base | Representation |
---|---|
bin | 100110111000001110111… |
… | …1011011011011000010011 |
3 | 200220211100121202122210011 |
4 | 1031300131323123120103 |
5 | 1200021332001343011 |
6 | 15210424302215351 |
7 | 1061023313533123 |
oct | 115603573333023 |
9 | 20824317678704 |
10 | 5343441434131 |
11 | 1780160428239 |
12 | 723717884557 |
13 | 2c9b65c47788 |
14 | 1468a416ac83 |
15 | 93eddaa6621 |
hex | 4dc1dedb613 |
5343441434131 has 4 divisors (see below), whose sum is σ = 5344429684672. Its totient is φ = 5342453183592.
The previous prime is 5343441434119. The next prime is 5343441434161. The reversal of 5343441434131 is 1314341443435.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 5343441434131 - 229 = 5342904563219 is a prime.
It is a super-2 number, since 2×53434414341312 (a number of 26 digits) contains 22 as substring.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (5343441434161) 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, 494117160 + ... + 494127973.
It is an arithmetic number, because the mean of its divisors is an integer number (1336107421168).
Almost surely, 25343441434131 is an apocalyptic number.
5343441434131 is a deficient number, since it is larger than the sum of its proper divisors (988250541).
5343441434131 is an equidigital number, since it uses as much as digits as its factorization.
5343441434131 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 988250540.
The product of its digits is 414720, while the sum is 40.
Adding to 5343441434131 its reverse (1314341443435), we get a palindrome (6657782877566).
The spelling of 5343441434131 in words is "five trillion, three hundred forty-three billion, four hundred forty-one million, four hundred thirty-four thousand, one hundred thirty-one".
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