Base | Representation |
---|---|
bin | 1010000000011110000111… |
… | …0101110001000010011011 |
3 | 1102221220011221102100110021 |
4 | 2200013201311301002123 |
5 | 2420234023223310042 |
6 | 35222444515131311 |
7 | 2213645362431652 |
oct | 240074165610233 |
9 | 42856157370407 |
10 | 11003200213147 |
11 | 356248173846a |
12 | 12985b15a3b37 |
13 | 61a79c4388bc |
14 | 2a07b45b8199 |
15 | 1413425ea867 |
hex | a01e1d7109b |
11003200213147 has 2 divisors, whose sum is σ = 11003200213148. Its totient is φ = 11003200213146.
The previous prime is 11003200213133. The next prime is 11003200213207. The reversal of 11003200213147 is 74131200230011.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 11003200213147 - 211 = 11003200211099 is a prime.
It is not a weakly prime, because it can be changed into another prime (11003000213147) 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, 5501600106573 + 5501600106574.
It is an arithmetic number, because the mean of its divisors is an integer number (5501600106574).
Almost surely, 211003200213147 is an apocalyptic number.
11003200213147 is a deficient number, since it is larger than the sum of its proper divisors (1).
11003200213147 is an equidigital number, since it uses as much as digits as its factorization.
11003200213147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1008, while the sum is 25.
Adding to 11003200213147 its reverse (74131200230011), we get a palindrome (85134400443158).
The spelling of 11003200213147 in words is "eleven trillion, three billion, two hundred million, two hundred thirteen thousand, one hundred forty-seven".
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