Base | Representation |
---|---|
bin | 1100000000011010011100… |
… | …1110111110001100100111 |
3 | 1201202000201010110221101211 |
4 | 3000012213032332030213 |
5 | 3212242114444023310 |
6 | 44024322132532251 |
7 | 2531521016300440 |
oct | 300064716761447 |
9 | 51660633427354 |
10 | 13201240220455 |
11 | 422a683a48981 |
12 | 15925a4781087 |
13 | 749b43c11a41 |
14 | 338d2cb062c7 |
15 | 17d5db93dc8a |
hex | c01a73be327 |
13201240220455 has 64 divisors (see below), whose sum is σ = 18446247936000. Its totient is φ = 8882033321088.
The previous prime is 13201240220437. The next prime is 13201240220471. The reversal of 13201240220455 is 55402204210231.
It is a happy number.
13201240220455 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a cyclic number.
It is not a de Polignac number, because 13201240220455 - 27 = 13201240220327 is a prime.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 3128995336 + ... + 3128999554.
It is an arithmetic number, because the mean of its divisors is an integer number (288222624000).
Almost surely, 213201240220455 is an apocalyptic number.
13201240220455 is a deficient number, since it is larger than the sum of its proper divisors (5245007715545).
13201240220455 is a wasteful number, since it uses less digits than its factorization.
13201240220455 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 6796.
The product of its (nonzero) digits is 19200, while the sum is 31.
Adding to 13201240220455 its reverse (55402204210231), we get a palindrome (68603444430686).
The spelling of 13201240220455 in words is "thirteen trillion, two hundred one billion, two hundred forty million, two hundred twenty thousand, four hundred fifty-five".
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