Base | Representation |
---|---|
bin | 1001001001110111000011… |
… | …01100100101101000101011 |
3 | 2122021102001112121222022122 |
4 | 10210323201230211220223 |
5 | 10114302221321424042 |
6 | 110451332412141455 |
7 | 4145226150646502 |
oct | 444734154455053 |
9 | 78242045558278 |
10 | 20130003311147 |
11 | 64610a5098954 |
12 | 23113b546788b |
13 | b30335a77712 |
14 | 4d8423d49839 |
15 | 24d9631507d2 |
hex | 124ee1b25a2b |
20130003311147 has 2 divisors, whose sum is σ = 20130003311148. Its totient is φ = 20130003311146.
The previous prime is 20130003311081. The next prime is 20130003311183. The reversal of 20130003311147 is 74111330003102.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 20130003311147 - 28 = 20130003310891 is a prime.
It is a super-3 number, since 3×201300033111473 (a number of 41 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (20130003311947) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10065001655573 + 10065001655574.
It is an arithmetic number, because the mean of its divisors is an integer number (10065001655574).
Almost surely, 220130003311147 is an apocalyptic number.
20130003311147 is a deficient number, since it is larger than the sum of its proper divisors (1).
20130003311147 is an equidigital number, since it uses as much as digits as its factorization.
20130003311147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1512, while the sum is 26.
Adding to 20130003311147 its reverse (74111330003102), we get a palindrome (94241333314249).
The spelling of 20130003311147 in words is "twenty trillion, one hundred thirty billion, three million, three hundred eleven thousand, one hundred forty-seven".
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