Base | Representation |
---|---|
bin | 101100100000110… |
… | …1011111000101001 |
3 | 10212002111110101112 |
4 | 1121001223320221 |
5 | 11024331122441 |
6 | 404113200105 |
7 | 52004340056 |
oct | 13101537051 |
9 | 3762443345 |
10 | 1493614121 |
11 | 6a711943a |
12 | 358260035 |
13 | 1aa59774c |
14 | 10252002d |
15 | 8b1d74eb |
hex | 5906be29 |
1493614121 has 2 divisors, whose sum is σ = 1493614122. Its totient is φ = 1493614120.
The previous prime is 1493614037. The next prime is 1493614123. The reversal of 1493614121 is 1214163941.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1389798400 + 103815721 = 37280^2 + 10189^2 .
It is an emirp because it is prime and its reverse (1214163941) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1493614121 - 214 = 1493597737 is a prime.
It is a Sophie Germain prime.
Together with 1493614123, it forms a pair of twin primes.
It is a Chen prime.
It is a Curzon number.
It is not a weakly prime, because it can be changed into another prime (1493614123) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 746807060 + 746807061.
It is an arithmetic number, because the mean of its divisors is an integer number (746807061).
Almost surely, 21493614121 is an apocalyptic number.
It is an amenable number.
1493614121 is a deficient number, since it is larger than the sum of its proper divisors (1).
1493614121 is an equidigital number, since it uses as much as digits as its factorization.
1493614121 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 5184, while the sum is 32.
The square root of 1493614121 is about 38647.3041880026. The cubic root of 1493614121 is about 1143.0874859395.
The spelling of 1493614121 in words is "one billion, four hundred ninety-three million, six hundred fourteen thousand, one hundred twenty-one".
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