Base | Representation |
---|---|
bin | 10011101001111110… |
… | …110001000100100001 |
3 | 2000110212120010210001 |
4 | 103221332301010201 |
5 | 321210441230213 |
6 | 13410134140001 |
7 | 1345004030335 |
oct | 235176610441 |
9 | 60425503701 |
10 | 21105414433 |
11 | 8a50502274 |
12 | 41101b1601 |
13 | 1cb4702c0c |
14 | 10430337c5 |
15 | 837d211dd |
hex | 4e9fb1121 |
21105414433 has 2 divisors, whose sum is σ = 21105414434. Its totient is φ = 21105414432.
The previous prime is 21105414403. The next prime is 21105414443. The reversal of 21105414433 is 33441450112.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 11545072704 + 9560341729 = 107448^2 + 97777^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-21105414433 is a prime.
It is a super-2 number, since 2×211054144332 (a number of 21 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (21105414403) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10552707216 + 10552707217.
It is an arithmetic number, because the mean of its divisors is an integer number (10552707217).
Almost surely, 221105414433 is an apocalyptic number.
It is an amenable number.
21105414433 is a deficient number, since it is larger than the sum of its proper divisors (1).
21105414433 is an equidigital number, since it uses as much as digits as its factorization.
21105414433 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5760, while the sum is 28.
Adding to 21105414433 its reverse (33441450112), we get a palindrome (54546864545).
The spelling of 21105414433 in words is "twenty-one billion, one hundred five million, four hundred fourteen thousand, four hundred thirty-three".
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