Base | Representation |
---|---|
bin | 111100101010100010… |
… | …001101010011101001 |
3 | 20020010120212100012202 |
4 | 330222202031103221 |
5 | 2031400324314123 |
6 | 45531331545545 |
7 | 4464116612066 |
oct | 745242152351 |
9 | 206116770182 |
10 | 65138119913 |
11 | 25696852659 |
12 | 1075a75b2b5 |
13 | 61b10b8429 |
14 | 321d029c6d |
15 | 1a6388da28 |
hex | f2a88d4e9 |
65138119913 has 2 divisors, whose sum is σ = 65138119914. Its totient is φ = 65138119912.
The previous prime is 65138119903. The next prime is 65138119927. The reversal of 65138119913 is 31991183156.
65138119913 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 48036803929 + 17101315984 = 219173^2 + 130772^2 .
It is a cyclic number.
It is not a de Polignac number, because 65138119913 - 24 = 65138119897 is a prime.
It is not a weakly prime, because it can be changed into another prime (65138119903) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32569059956 + 32569059957.
It is an arithmetic number, because the mean of its divisors is an integer number (32569059957).
Almost surely, 265138119913 is an apocalyptic number.
It is an amenable number.
65138119913 is a deficient number, since it is larger than the sum of its proper divisors (1).
65138119913 is an equidigital number, since it uses as much as digits as its factorization.
65138119913 is an evil number, because the sum of its binary digits is even.
The product of its digits is 174960, while the sum is 47.
The spelling of 65138119913 in words is "sixty-five billion, one hundred thirty-eight million, one hundred nineteen thousand, nine hundred thirteen".
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