Base | Representation |
---|---|
bin | 11101001100110101000… |
… | …11110000000011101001 |
3 | 10112220202010102102010121 |
4 | 32212122203300003221 |
5 | 112414300013134423 |
6 | 2044530303043241 |
7 | 132326134024132 |
oct | 16463243600351 |
9 | 3486663372117 |
10 | 1003320443113 |
11 | 357562262a5a |
12 | 142549ba0521 |
13 | 737c7187b91 |
14 | 367bd424c89 |
15 | 1b17300645d |
hex | e99a8f00e9 |
1003320443113 has 2 divisors, whose sum is σ = 1003320443114. Its totient is φ = 1003320443112.
The previous prime is 1003320443039. The next prime is 1003320443173. The reversal of 1003320443113 is 3113440233001.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 742513102864 + 260807340249 = 861692^2 + 510693^2 .
It is a cyclic number.
It is not a de Polignac number, because 1003320443113 - 233 = 994730508521 is a prime.
It is not a weakly prime, because it can be changed into another prime (1003320443173) 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, 501660221556 + 501660221557.
It is an arithmetic number, because the mean of its divisors is an integer number (501660221557).
Almost surely, 21003320443113 is an apocalyptic number.
It is an amenable number.
1003320443113 is a deficient number, since it is larger than the sum of its proper divisors (1).
1003320443113 is an equidigital number, since it uses as much as digits as its factorization.
1003320443113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2592, while the sum is 25.
Adding to 1003320443113 its reverse (3113440233001), we get a palindrome (4116760676114).
The spelling of 1003320443113 in words is "one trillion, three billion, three hundred twenty million, four hundred forty-three thousand, one hundred thirteen".
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