Base | Representation |
---|---|
bin | 100101001000111110000… |
… | …1010000001110001001001 |
3 | 200001222121200010001102111 |
4 | 1022101330022001301021 |
5 | 1132113011030313423 |
6 | 14504550555102321 |
7 | 1034534241250423 |
oct | 112217412016111 |
9 | 20058550101374 |
10 | 5104504151113 |
11 | 1698898578aa4 |
12 | 6a53541419a1 |
13 | 2b0477aa1a83 |
14 | 1390b8ab3813 |
15 | 8cba70b960d |
hex | 4a47c281c49 |
5104504151113 has 2 divisors, whose sum is σ = 5104504151114. Its totient is φ = 5104504151112.
The previous prime is 5104504151041. The next prime is 5104504151137. The reversal of 5104504151113 is 3111514054015.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4729963823104 + 374540328009 = 2174848^2 + 611997^2 .
It is a cyclic number.
It is not a de Polignac number, because 5104504151113 - 221 = 5104502053961 is a prime.
It is not a weakly prime, because it can be changed into another prime (5104504151413) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2552252075556 + 2552252075557.
It is an arithmetic number, because the mean of its divisors is an integer number (2552252075557).
Almost surely, 25104504151113 is an apocalyptic number.
It is an amenable number.
5104504151113 is a deficient number, since it is larger than the sum of its proper divisors (1).
5104504151113 is an equidigital number, since it uses as much as digits as its factorization.
5104504151113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 6000, while the sum is 31.
The spelling of 5104504151113 in words is "five trillion, one hundred four billion, five hundred four million, one hundred fifty-one thousand, one hundred thirteen".
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