Base | Representation |
---|---|
bin | 11101100100010111001001… |
… | …110101010000110111100110 |
3 | 122001102220102022021221112121 |
4 | 131210113021311100313212 |
5 | 114021102220404113014 |
6 | 1140324304314540154 |
7 | 36251142511300330 |
oct | 3544271165206746 |
9 | 561386368257477 |
10 | 130042111004134 |
11 | 3848760551a98a |
12 | 12703047a3765a |
13 | 5773bc2aa9204 |
14 | 2418100c7c450 |
15 | 1007a6156d124 |
hex | 7645c9d50de6 |
130042111004134 has 8 divisors (see below), whose sum is σ = 222929333149968. Its totient is φ = 55732333287480.
The previous prime is 130042111004129. The next prime is 130042111004147. The reversal of 130042111004134 is 431400111240031.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a junction number, because it is equal to n+sod(n) for n = 130042111004099 and 130042111004108.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 4644361107277 + ... + 4644361107304.
It is an arithmetic number, because the mean of its divisors is an integer number (27866166643746).
Almost surely, 2130042111004134 is an apocalyptic number.
130042111004134 is a gapful number since it is divisible by the number (14) formed by its first and last digit.
130042111004134 is a deficient number, since it is larger than the sum of its proper divisors (92887222145834).
130042111004134 is an equidigital number, since it uses as much as digits as its factorization.
130042111004134 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 9288722214590.
The product of its (nonzero) digits is 1152, while the sum is 25.
Adding to 130042111004134 its reverse (431400111240031), we get a palindrome (561442222244165).
The spelling of 130042111004134 in words is "one hundred thirty trillion, forty-two billion, one hundred eleven million, four thousand, one hundred thirty-four".
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