Base | Representation |
---|---|
bin | 10010111111100110011… |
… | …100000010010111110001 |
3 | 11121210001111121000010212 |
4 | 102333212130002113301 |
5 | 132341113221400423 |
6 | 2435341452433505 |
7 | 163205044100234 |
oct | 22774634022761 |
9 | 4553044530125 |
10 | 1305241200113 |
11 | 4636056a2434 |
12 | 190b6a778895 |
13 | 96111c07196 |
14 | 472616b541b |
15 | 23e44123578 |
hex | 12fe67025f1 |
1305241200113 has 2 divisors, whose sum is σ = 1305241200114. Its totient is φ = 1305241200112.
The previous prime is 1305241200041. The next prime is 1305241200139. The reversal of 1305241200113 is 3110021425031.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 848255736064 + 456985464049 = 921008^2 + 676007^2 .
It is a cyclic number.
It is not a de Polignac number, because 1305241200113 - 224 = 1305224422897 is a prime.
It is a super-2 number, since 2×13052412001132 (a number of 25 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (1305241200013) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 652620600056 + 652620600057.
It is an arithmetic number, because the mean of its divisors is an integer number (652620600057).
Almost surely, 21305241200113 is an apocalyptic number.
It is an amenable number.
1305241200113 is a deficient number, since it is larger than the sum of its proper divisors (1).
1305241200113 is an equidigital number, since it uses as much as digits as its factorization.
1305241200113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 720, while the sum is 23.
Adding to 1305241200113 its reverse (3110021425031), we get a palindrome (4415262625144).
The spelling of 1305241200113 in words is "one trillion, three hundred five billion, two hundred forty-one million, two hundred thousand, one hundred thirteen".
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