Base | Representation |
---|---|
bin | 1100000010110000001001… |
… | …0000100000011101000011 |
3 | 1201212212102110200122022210 |
4 | 3000230002100200131003 |
5 | 3213421413021231420 |
6 | 44055005245033203 |
7 | 2534442532251003 |
oct | 300540220403503 |
9 | 51785373618283 |
10 | 13241422055235 |
11 | 42457236020a0 |
12 | 159a339555803 |
13 | 75087885b33a |
14 | 33ac614c8003 |
15 | 17e68e3929e0 |
hex | c0b02420743 |
13241422055235 has 32 divisors (see below), whose sum is σ = 23127221548800. Its totient is φ = 6415938633600.
The previous prime is 13241422055227. The next prime is 13241422055333. The reversal of 13241422055235 is 53255022414231.
It is not a de Polignac number, because 13241422055235 - 23 = 13241422055227 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13241422055193 and 13241422055202.
It is an unprimeable number.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 25648561 + ... + 26159730.
It is an arithmetic number, because the mean of its divisors is an integer number (722725673400).
Almost surely, 213241422055235 is an apocalyptic number.
13241422055235 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
13241422055235 is a deficient number, since it is larger than the sum of its proper divisors (9885799493565).
13241422055235 is a wasteful number, since it uses less digits than its factorization.
13241422055235 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 51809859.
The product of its (nonzero) digits is 288000, while the sum is 39.
Adding to 13241422055235 its reverse (53255022414231), we get a palindrome (66496444469466).
The spelling of 13241422055235 in words is "thirteen trillion, two hundred forty-one billion, four hundred twenty-two million, fifty-five thousand, two hundred thirty-five".
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