Base | Representation |
---|---|
bin | 1010010011101111010110… |
… | …0111001010100001001111 |
3 | 1111010112122222111001101111 |
4 | 2210323311213022201033 |
5 | 2441200013032131341 |
6 | 40034514025152451 |
7 | 2246605066616212 |
oct | 244736547124117 |
9 | 44115588431344 |
10 | 11334244411471 |
11 | 367a90a53a413 |
12 | 133079b62b727 |
13 | 642a78a73217 |
14 | 2b281a6d9779 |
15 | 149c6a3e5681 |
hex | a4ef59ca84f |
11334244411471 has 2 divisors, whose sum is σ = 11334244411472. Its totient is φ = 11334244411470.
The previous prime is 11334244411421. The next prime is 11334244411493. The reversal of 11334244411471 is 17411444243311.
It is a strong prime.
It is an emirp because it is prime and its reverse (17411444243311) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 11334244411471 - 215 = 11334244378703 is a prime.
It is a super-3 number, since 3×113342444114713 (a number of 40 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11334244411421) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5667122205735 + 5667122205736.
It is an arithmetic number, because the mean of its divisors is an integer number (5667122205736).
Almost surely, 211334244411471 is an apocalyptic number.
11334244411471 is a deficient number, since it is larger than the sum of its proper divisors (1).
11334244411471 is an equidigital number, since it uses as much as digits as its factorization.
11334244411471 is an evil number, because the sum of its binary digits is even.
The product of its digits is 129024, while the sum is 40.
Adding to 11334244411471 its reverse (17411444243311), we get a palindrome (28745688654782).
The spelling of 11334244411471 in words is "eleven trillion, three hundred thirty-four billion, two hundred forty-four million, four hundred eleven thousand, four hundred seventy-one".
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