Base | Representation |
---|---|
bin | 1100000110001110100000… |
… | …1001011000000001001011 |
3 | 1202002120111110201020120121 |
4 | 3001203220021120001023 |
5 | 3220411140030121303 |
6 | 44142240332204111 |
7 | 2541654664320445 |
oct | 301435011300113 |
9 | 52076443636517 |
10 | 13301113520203 |
11 | 4268a74906223 |
12 | 15a9a18009637 |
13 | 75639c3bc902 |
14 | 33dac4dc5495 |
15 | 180ed499c9bd |
hex | c18e825804b |
13301113520203 has 2 divisors, whose sum is σ = 13301113520204. Its totient is φ = 13301113520202.
The previous prime is 13301113520161. The next prime is 13301113520221. The reversal of 13301113520203 is 30202531110331.
It is a strong prime.
It is an emirp because it is prime and its reverse (30202531110331) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 13301113520203 - 217 = 13301113389131 is a prime.
It is a super-2 number, since 2×133011135202032 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (13301113520803) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6650556760101 + 6650556760102.
It is an arithmetic number, because the mean of its divisors is an integer number (6650556760102).
Almost surely, 213301113520203 is an apocalyptic number.
13301113520203 is a deficient number, since it is larger than the sum of its proper divisors (1).
13301113520203 is an equidigital number, since it uses as much as digits as its factorization.
13301113520203 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1620, while the sum is 25.
Adding to 13301113520203 its reverse (30202531110331), we get a palindrome (43503644630534).
The spelling of 13301113520203 in words is "thirteen trillion, three hundred one billion, one hundred thirteen million, five hundred twenty thousand, two hundred three".
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