Base | Representation |
---|---|
bin | 10000001001110001001… |
… | …00111110111111101101 |
3 | 1222001112220102020002211 |
4 | 20010320210332333231 |
5 | 33043120000003433 |
6 | 1102544040050421 |
7 | 55045263111121 |
oct | 10047044767755 |
9 | 1861486366084 |
10 | 555000000493 |
11 | 1a4413038964 |
12 | 8b690463a11 |
13 | 4044aa49673 |
14 | 1cc0da15581 |
15 | e68442e8cd |
hex | 813893efed |
555000000493 has 2 divisors, whose sum is σ = 555000000494. Its totient is φ = 555000000492.
The previous prime is 555000000491. The next prime is 555000000547. The reversal of 555000000493 is 394000000555.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 308410290409 + 246589710084 = 555347^2 + 496578^2 .
It is a cyclic number.
It is not a de Polignac number, because 555000000493 - 21 = 555000000491 is a prime.
Together with 555000000491, it forms a pair of twin primes.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (555000000491) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 277500000246 + 277500000247.
It is an arithmetic number, because the mean of its divisors is an integer number (277500000247).
Almost surely, 2555000000493 is an apocalyptic number.
It is an amenable number.
555000000493 is a deficient number, since it is larger than the sum of its proper divisors (1).
555000000493 is an equidigital number, since it uses as much as digits as its factorization.
555000000493 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 13500, while the sum is 31.
The spelling of 555000000493 in words is "five hundred fifty-five billion, four hundred ninety-three", and thus it is an aban number.
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