Base | Representation |
---|---|
bin | 10111101101100111101101… |
… | …10101010000111001001011 |
3 | 20211122000110210010102022201 |
4 | 23312303312311100321023 |
5 | 23313321022244201103 |
6 | 302523034053301031 |
7 | 13661232542616016 |
oct | 1366636665207113 |
9 | 224560423112281 |
10 | 52145044131403 |
11 | 15684652aa2653 |
12 | 5a22085934777 |
13 | 231334488a072 |
14 | cc3b9559437d |
15 | 60662b5d151d |
hex | 2f6cf6d50e4b |
52145044131403 has 4 divisors (see below), whose sum is σ = 52257668417248. Its totient is φ = 52032419845560.
The previous prime is 52145044131289. The next prime is 52145044131419. The reversal of 52145044131403 is 30413144054125.
It is a happy number.
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 52145044131403 - 221 = 52145042034251 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (52145044131203) 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, 56312142228 + ... + 56312143153.
It is an arithmetic number, because the mean of its divisors is an integer number (13064417104312).
Almost surely, 252145044131403 is an apocalyptic number.
52145044131403 is a deficient number, since it is larger than the sum of its proper divisors (112624285845).
52145044131403 is a wasteful number, since it uses less digits than its factorization.
52145044131403 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 112624285844.
The product of its (nonzero) digits is 115200, while the sum is 37.
Adding to 52145044131403 its reverse (30413144054125), we get a palindrome (82558188185528).
The spelling of 52145044131403 in words is "fifty-two trillion, one hundred forty-five billion, forty-four million, one hundred thirty-one thousand, four hundred three".
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