Base | Representation |
---|---|
bin | 11110011110011001101010… |
… | …111111010111001110001111 |
3 | 122120120012021002110000012122 |
4 | 132132121222333113032033 |
5 | 120031424023333342223 |
6 | 1153020440533123155 |
7 | 40142246560134014 |
oct | 3636315277271617 |
9 | 576505232400178 |
10 | 134030544434063 |
11 | 39785042189232 |
12 | 130480261894bb |
13 | 59a3045487934 |
14 | 2515181d8770b |
15 | 1076696d715c8 |
hex | 79e66afd738f |
134030544434063 has 2 divisors, whose sum is σ = 134030544434064. Its totient is φ = 134030544434062.
The previous prime is 134030544434023. The next prime is 134030544434069. The reversal of 134030544434063 is 360434445030431.
It is a strong prime.
It is an emirp because it is prime and its reverse (360434445030431) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 134030544434063 - 230 = 134029470692239 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (134030544434069) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (31) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67015272217031 + 67015272217032.
It is an arithmetic number, because the mean of its divisors is an integer number (67015272217032).
Almost surely, 2134030544434063 is an apocalyptic number.
134030544434063 is a deficient number, since it is larger than the sum of its proper divisors (1).
134030544434063 is an equidigital number, since it uses as much as digits as its factorization.
134030544434063 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2488320, while the sum is 44.
Adding to 134030544434063 its reverse (360434445030431), we get a palindrome (494464989464494).
The spelling of 134030544434063 in words is "one hundred thirty-four trillion, thirty billion, five hundred forty-four million, four hundred thirty-four thousand, sixty-three".
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