Base | Representation |
---|---|
bin | 100110100101001011001… |
… | …0011000011011011010001 |
3 | 200202220201010100100222111 |
4 | 1031022112103003123101 |
5 | 1143334020333023213 |
6 | 15135535013505321 |
7 | 1055044112465050 |
oct | 115122623033321 |
9 | 20686633310874 |
10 | 5302511220433 |
11 | 1764867390949 |
12 | 7177b4342241 |
13 | 2c604225050b |
14 | 14490021a397 |
15 | 92de55c5a3d |
hex | 4d2964c36d1 |
5302511220433 has 8 divisors (see below), whose sum is σ = 6060036423168. Its totient is φ = 4544991917664.
The previous prime is 5302511220379. The next prime is 5302511220437. The reversal of 5302511220433 is 3340221152035.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 5302511220433 - 221 = 5302509123281 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 5302511220395 and 5302511220404.
It is not an unprimeable number, because it can be changed into a prime (5302511220437) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 656160 + ... + 3321982.
It is an arithmetic number, because the mean of its divisors is an integer number (757504552896).
Almost surely, 25302511220433 is an apocalyptic number.
It is an amenable number.
5302511220433 is a deficient number, since it is larger than the sum of its proper divisors (757525202735).
5302511220433 is a wasteful number, since it uses less digits than its factorization.
5302511220433 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2949983.
The product of its (nonzero) digits is 21600, while the sum is 31.
Adding to 5302511220433 its reverse (3340221152035), we get a palindrome (8642732372468).
The spelling of 5302511220433 in words is "five trillion, three hundred two billion, five hundred eleven million, two hundred twenty thousand, four hundred thirty-three".
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