Base | Representation |
---|---|
bin | 10011100100100001110010… |
… | …10101000101010101110011 |
3 | 12122101020211001121121002112 |
4 | 21302100321111011111303 |
5 | 21120102310214243011 |
6 | 231310413313035535 |
7 | 12031200302236316 |
oct | 1162207125052563 |
9 | 178336731547075 |
10 | 43036534134131 |
11 | 1279276761652a |
12 | 49b0928255bab |
13 | 1b024351021c5 |
14 | a8ad9d54897d |
15 | 4e972ba2e58b |
hex | 272439545573 |
43036534134131 has 2 divisors, whose sum is σ = 43036534134132. Its totient is φ = 43036534134130.
The previous prime is 43036534134109. The next prime is 43036534134197. The reversal of 43036534134131 is 13143143563034.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 43036534134131 - 230 = 43035460392307 is a prime.
It is a super-2 number, since 2×430365341341312 (a number of 28 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (43036534134631) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 21518267067065 + 21518267067066.
It is an arithmetic number, because the mean of its divisors is an integer number (21518267067066).
Almost surely, 243036534134131 is an apocalyptic number.
43036534134131 is a deficient number, since it is larger than the sum of its proper divisors (1).
43036534134131 is an equidigital number, since it uses as much as digits as its factorization.
43036534134131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 466560, while the sum is 41.
Adding to 43036534134131 its reverse (13143143563034), we get a palindrome (56179677697165).
The spelling of 43036534134131 in words is "forty-three trillion, thirty-six billion, five hundred thirty-four million, one hundred thirty-four thousand, one hundred thirty-one".
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