Base | Representation |
---|---|
bin | 1100000000011001111011… |
… | …1111101101101111001011 |
3 | 1201202000100111101202220021 |
4 | 3000012132333231233023 |
5 | 3212241324103100042 |
6 | 44024300314241311 |
7 | 2531514420413512 |
oct | 300063677555713 |
9 | 51660314352807 |
10 | 13201102003147 |
11 | 422a612a241a3 |
12 | 1592566426237 |
13 | 749b213a9c56 |
14 | 338d18607679 |
15 | 17d5ce73a867 |
hex | c019efedbcb |
13201102003147 has 2 divisors, whose sum is σ = 13201102003148. Its totient is φ = 13201102003146.
The previous prime is 13201102003109. The next prime is 13201102003169. The reversal of 13201102003147 is 74130020110231.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13201102003147 - 235 = 13166742264779 is a prime.
It is not a weakly prime, because it can be changed into another prime (13201102003547) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6600551001573 + 6600551001574.
It is an arithmetic number, because the mean of its divisors is an integer number (6600551001574).
Almost surely, 213201102003147 is an apocalyptic number.
13201102003147 is a deficient number, since it is larger than the sum of its proper divisors (1).
13201102003147 is an equidigital number, since it uses as much as digits as its factorization.
13201102003147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1008, while the sum is 25.
Adding to 13201102003147 its reverse (74130020110231), we get a palindrome (87331122113378).
The spelling of 13201102003147 in words is "thirteen trillion, two hundred one billion, one hundred two million, three thousand, one hundred forty-seven".
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