Base | Representation |
---|---|
bin | 100101011001100010111… |
… | …1010111010001001110001 |
3 | 200012101111111200200211121 |
4 | 1022303011322322021301 |
5 | 1133203411432040423 |
6 | 14533155140342241 |
7 | 1040234335232203 |
oct | 112630572721161 |
9 | 20171444620747 |
10 | 5140101440113 |
11 | 17019a5296102 |
12 | 6b02296a4981 |
13 | 2b392b9823a3 |
14 | 13aad465c773 |
15 | 8da8c2ce45d |
hex | 4acc5eba271 |
5140101440113 has 2 divisors, whose sum is σ = 5140101440114. Its totient is φ = 5140101440112.
The previous prime is 5140101440081. The next prime is 5140101440143. The reversal of 5140101440113 is 3110441010415.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4177805185024 + 962296255089 = 2043968^2 + 980967^2 .
It is a cyclic number.
It is not a de Polignac number, because 5140101440113 - 25 = 5140101440081 is a prime.
It is not a weakly prime, because it can be changed into another prime (5140101440143) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2570050720056 + 2570050720057.
It is an arithmetic number, because the mean of its divisors is an integer number (2570050720057).
Almost surely, 25140101440113 is an apocalyptic number.
It is an amenable number.
5140101440113 is a deficient number, since it is larger than the sum of its proper divisors (1).
5140101440113 is an equidigital number, since it uses as much as digits as its factorization.
5140101440113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 960, while the sum is 25.
Adding to 5140101440113 its reverse (3110441010415), we get a palindrome (8250542450528).
The spelling of 5140101440113 in words is "five trillion, one hundred forty billion, one hundred one million, four hundred forty thousand, one hundred thirteen".
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