Base | Representation |
---|---|
bin | 100100101000010101011… |
… | …0110001010010101110001 |
3 | 122211021201102101022102011 |
4 | 1021100222312022111301 |
5 | 1124440443122341423 |
6 | 14412440355322521 |
7 | 1026503430200326 |
oct | 111205266122561 |
9 | 18737642338364 |
10 | 5034420512113 |
11 | 16710a414a793 |
12 | 6938552b0441 |
13 | 2a69892218b5 |
14 | 13594ad28b4d |
15 | 8ae54486b0d |
hex | 4942ad8a571 |
5034420512113 has 2 divisors, whose sum is σ = 5034420512114. Its totient is φ = 5034420512112.
The previous prime is 5034420512083. The next prime is 5034420512171. The reversal of 5034420512113 is 3112150244305.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 3763103376384 + 1271317135729 = 1939872^2 + 1127527^2 .
It is a cyclic number.
It is not a de Polignac number, because 5034420512113 - 213 = 5034420503921 is a prime.
It is not a weakly prime, because it can be changed into another prime (5034420512413) 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, 2517210256056 + 2517210256057.
It is an arithmetic number, because the mean of its divisors is an integer number (2517210256057).
Almost surely, 25034420512113 is an apocalyptic number.
It is an amenable number.
5034420512113 is a deficient number, since it is larger than the sum of its proper divisors (1).
5034420512113 is an equidigital number, since it uses as much as digits as its factorization.
5034420512113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 14400, while the sum is 31.
Adding to 5034420512113 its reverse (3112150244305), we get a palindrome (8146570756418).
The spelling of 5034420512113 in words is "five trillion, thirty-four billion, four hundred twenty million, five hundred twelve thousand, one hundred thirteen".
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