Base | Representation |
---|---|
bin | 1011101110010010001… |
… | …1110111000100101001 |
3 | 201020212010110121212221 |
4 | 2323210203313010221 |
5 | 11244433143221411 |
6 | 232305011234041 |
7 | 20356646054611 |
oct | 2734443670451 |
9 | 636763417787 |
10 | 201403101481 |
11 | 784618890aa |
12 | 330495a8921 |
13 | 15cb8c1553a |
14 | 9a6859da41 |
15 | 538b72d071 |
hex | 2ee48f7129 |
201403101481 has 2 divisors, whose sum is σ = 201403101482. Its totient is φ = 201403101480.
The previous prime is 201403101469. The next prime is 201403101491. The reversal of 201403101481 is 184101304102.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 117022147225 + 84380954256 = 342085^2 + 290484^2 .
It is a cyclic number.
It is not a de Polignac number, because 201403101481 - 211 = 201403099433 is a prime.
It is a super-2 number, since 2×2014031014812 (a number of 23 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (201403101491) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100701550740 + 100701550741.
It is an arithmetic number, because the mean of its divisors is an integer number (100701550741).
Almost surely, 2201403101481 is an apocalyptic number.
It is an amenable number.
201403101481 is a deficient number, since it is larger than the sum of its proper divisors (1).
201403101481 is an equidigital number, since it uses as much as digits as its factorization.
201403101481 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 768, while the sum is 25.
Adding to 201403101481 its reverse (184101304102), we get a palindrome (385504405583).
The spelling of 201403101481 in words is "two hundred one billion, four hundred three million, one hundred one thousand, four hundred eighty-one".
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