Base | Representation |
---|---|
bin | 100110101001010001111… |
… | …1101010010100111111111 |
3 | 200210202110202001121112012 |
4 | 1031102203331102213333 |
5 | 1144010101031044020 |
6 | 15143554054001435 |
7 | 1055505500625053 |
oct | 115224375224777 |
9 | 20722422047465 |
10 | 5311330331135 |
11 | 1768582562294 |
12 | 71945594627b |
13 | 2c6b193a7c0a |
14 | 1450d95c5c63 |
15 | 9325e9707c5 |
hex | 4d4a3f529ff |
5311330331135 has 16 divisors (see below), whose sum is σ = 6478352358912. Its totient is φ = 4179232780800.
The previous prime is 5311330331129. The next prime is 5311330331171.
5311330331135 is nontrivially palindromic in base 10.
5311330331135 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 5311330331135 - 214 = 5311330314751 is a prime.
It is a super-3 number, since 3×53113303311353 (a number of 39 digits) contains 333 as substring.
It is a Duffinian number.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 7202375 + ... + 7905495.
It is an arithmetic number, because the mean of its divisors is an integer number (404897022432).
Almost surely, 25311330331135 is an apocalyptic number.
5311330331135 is a deficient number, since it is larger than the sum of its proper divisors (1167022027777).
5311330331135 is a wasteful number, since it uses less digits than its factorization.
5311330331135 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 727954.
The product of its (nonzero) digits is 18225, while the sum is 32.
It can be divided in two parts, 5311330 and 331135, that added together give a palindrome (5642465).
The spelling of 5311330331135 in words is "five trillion, three hundred eleven billion, three hundred thirty million, three hundred thirty-one thousand, one hundred thirty-five".
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