Base | Representation |
---|---|
bin | 11101010010111000000… |
… | …110001101101100000011 |
3 | 21010110020201112012110021 |
4 | 131102320012031230003 |
5 | 230440344113044303 |
6 | 4140453133514311 |
7 | 265305214656301 |
oct | 35227006155403 |
9 | 7113221465407 |
10 | 2013133331203 |
11 | 706846312a15 |
12 | 2861aa342997 |
13 | 117ab7449b77 |
14 | 6d616c05271 |
15 | 37575eb9ebd |
hex | 1d4b818db03 |
2013133331203 has 2 divisors, whose sum is σ = 2013133331204. Its totient is φ = 2013133331202.
The previous prime is 2013133331173. The next prime is 2013133331213. The reversal of 2013133331203 is 3021333313102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 2013133331203 - 25 = 2013133331171 is a prime.
It is a super-4 number, since 4×20131333312034 (a number of 50 digits) contains 4444 as substring.
It is not a weakly prime, because it can be changed into another prime (2013133331213) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1006566665601 + 1006566665602.
It is an arithmetic number, because the mean of its divisors is an integer number (1006566665602).
Almost surely, 22013133331203 is an apocalyptic number.
2013133331203 is a deficient number, since it is larger than the sum of its proper divisors (1).
2013133331203 is an equidigital number, since it uses as much as digits as its factorization.
2013133331203 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2916, while the sum is 25.
Adding to 2013133331203 its reverse (3021333313102), we get a palindrome (5034466644305).
The spelling of 2013133331203 in words is "two trillion, thirteen billion, one hundred thirty-three million, three hundred thirty-one thousand, two hundred three".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.062 sec. • engine limits •