Base | Representation |
---|---|
bin | 1111110101111010110… |
… | …01011011101011111111 |
3 | 1221001001020012212110212 |
4 | 13322331121123223333 |
5 | 32404303410041002 |
6 | 1054022401023035 |
7 | 54220223620034 |
oct | 7727531335377 |
9 | 1831036185425 |
10 | 544343440127 |
11 | 19a944754383 |
12 | 895b7624a7b |
13 | 3c44001c146 |
14 | 1c4bc5c538b |
15 | e25dad0352 |
hex | 7ebd65baff |
544343440127 has 2 divisors, whose sum is σ = 544343440128. Its totient is φ = 544343440126.
The previous prime is 544343440061. The next prime is 544343440129. The reversal of 544343440127 is 721044343445.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 544343440127 - 210 = 544343439103 is a prime.
It is a super-2 number, since 2×5443434401272 (a number of 24 digits) contains 22 as substring.
Together with 544343440129, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (544343440129) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 272171720063 + 272171720064.
It is an arithmetic number, because the mean of its divisors is an integer number (272171720064).
Almost surely, 2544343440127 is an apocalyptic number.
544343440127 is a deficient number, since it is larger than the sum of its proper divisors (1).
544343440127 is an equidigital number, since it uses as much as digits as its factorization.
544343440127 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 645120, while the sum is 41.
The spelling of 544343440127 in words is "five hundred forty-four billion, three hundred forty-three million, four hundred forty thousand, one hundred twenty-seven".
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