Base | Representation |
---|---|
bin | 1011111011000110111101… |
… | …1110010111110011101011 |
3 | 1201102022111010200102200201 |
4 | 2332301233132113303223 |
5 | 3204244001231421303 |
6 | 43514411313013031 |
7 | 2522113520623546 |
oct | 276615736276353 |
9 | 51368433612621 |
10 | 13110110420203 |
11 | 41a4a6a609175 |
12 | 15789b1554177 |
13 | 741380006a04 |
14 | 334765b4d85d |
15 | 17b0562e181d |
hex | bec6f797ceb |
13110110420203 has 2 divisors, whose sum is σ = 13110110420204. Its totient is φ = 13110110420202.
The previous prime is 13110110420201. The next prime is 13110110420221. The reversal of 13110110420203 is 30202401101131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13110110420203 - 21 = 13110110420201 is a prime.
It is a super-2 number, since 2×131101104202032 (a number of 27 digits) contains 22 as substring.
Together with 13110110420201, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (13110110420201) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6555055210101 + 6555055210102.
It is an arithmetic number, because the mean of its divisors is an integer number (6555055210102).
Almost surely, 213110110420203 is an apocalyptic number.
13110110420203 is a deficient number, since it is larger than the sum of its proper divisors (1).
13110110420203 is an equidigital number, since it uses as much as digits as its factorization.
13110110420203 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 144, while the sum is 19.
Adding to 13110110420203 its reverse (30202401101131), we get a palindrome (43312511521334).
The spelling of 13110110420203 in words is "thirteen trillion, one hundred ten billion, one hundred ten million, four hundred twenty thousand, two hundred three".
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