Base | Representation |
---|---|
bin | 100110011111011101… |
… | …110101010010111001 |
3 | 10221200100212210110101 |
4 | 212133131311102321 |
5 | 1134121001302411 |
6 | 30553051315401 |
7 | 2662125040342 |
oct | 463735652271 |
9 | 127610783411 |
10 | 41330103481 |
11 | 16589847517 |
12 | 801544bb61 |
13 | 3b887ca644 |
14 | 20010c70c9 |
15 | 111d6718c1 |
hex | 99f7754b9 |
41330103481 has 2 divisors, whose sum is σ = 41330103482. Its totient is φ = 41330103480.
The previous prime is 41330103439. The next prime is 41330103487. The reversal of 41330103481 is 18430103314.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 39158077456 + 2172026025 = 197884^2 + 46605^2 .
It is a cyclic number.
It is not a de Polignac number, because 41330103481 - 223 = 41321714873 is a prime.
It is a super-3 number, since 3×413301034813 (a number of 33 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not a weakly prime, because it can be changed into another prime (41330103487) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20665051740 + 20665051741.
It is an arithmetic number, because the mean of its divisors is an integer number (20665051741).
Almost surely, 241330103481 is an apocalyptic number.
It is an amenable number.
41330103481 is a deficient number, since it is larger than the sum of its proper divisors (1).
41330103481 is an equidigital number, since it uses as much as digits as its factorization.
41330103481 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3456, while the sum is 28.
Adding to 41330103481 its reverse (18430103314), we get a palindrome (59760206795).
The spelling of 41330103481 in words is "forty-one billion, three hundred thirty million, one hundred three thousand, four hundred eighty-one".
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