Base | Representation |
---|---|
bin | 10101000110101001… |
… | …00001110111101010 |
3 | 1002020121101202102220 |
4 | 22203110201313222 |
5 | 141200440034120 |
6 | 5112133134510 |
7 | 550524300666 |
oct | 124324416752 |
9 | 32217352386 |
10 | 11330002410 |
11 | 4894544561 |
12 | 224248a436 |
13 | 10b73cb54c |
14 | 796a538a6 |
15 | 464a27b40 |
hex | 2a3521dea |
11330002410 has 32 divisors (see below), whose sum is σ = 28791536832. Its totient is φ = 2843608320.
The previous prime is 11330002381. The next prime is 11330002441. The reversal of 11330002410 is 1420003311.
11330002410 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a super-2 number, since 2×113300024102 (a number of 21 digits) contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (15).
It is a self number, because there is not a number n which added to its sum of digits gives 11330002410.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 11107336 + ... + 11108355.
It is an arithmetic number, because the mean of its divisors is an integer number (899735526).
Almost surely, 211330002410 is an apocalyptic number.
11330002410 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
11330002410 is an abundant number, since it is smaller than the sum of its proper divisors (17461534422).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
11330002410 is a wasteful number, since it uses less digits than its factorization.
11330002410 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 22215718.
The product of its (nonzero) digits is 72, while the sum is 15.
Adding to 11330002410 its reverse (1420003311), we get a palindrome (12750005721).
Subtracting from 11330002410 its reverse (1420003311), we obtain a palindrome (9909999099).
The spelling of 11330002410 in words is "eleven billion, three hundred thirty million, two thousand, four hundred ten".
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