Base | Representation |
---|---|
bin | 100000001111110100100… |
… | …0110010000010011000001 |
3 | 120200200211020022212211221 |
4 | 1000133221012100103001 |
5 | 1040103240113404441 |
6 | 13232013125035041 |
7 | 635126431350013 |
oct | 100375106202301 |
9 | 16620736285757 |
10 | 4432022013121 |
11 | 145967897a41a |
12 | 5b6b57255a81 |
13 | 261c266b2a21 |
14 | 1147222d61b3 |
15 | 7a448c8c8d1 |
hex | 407e91904c1 |
4432022013121 has 32 divisors (see below), whose sum is σ = 4681762630272. Its totient is φ = 4190011338240.
The previous prime is 4432022013073. The next prime is 4432022013143. The reversal of 4432022013121 is 1213102202344.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-4432022013121 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (4432022013151) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 387510415 + ... + 387521851.
It is an arithmetic number, because the mean of its divisors is an integer number (146305082196).
Almost surely, 24432022013121 is an apocalyptic number.
4432022013121 is a gapful number since it is divisible by the number (41) formed by its first and last digit.
It is an amenable number.
4432022013121 is a deficient number, since it is larger than the sum of its proper divisors (249740617151).
4432022013121 is a wasteful number, since it uses less digits than its factorization.
4432022013121 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 12529.
The product of its (nonzero) digits is 2304, while the sum is 25.
Adding to 4432022013121 its reverse (1213102202344), we get a palindrome (5645124215465).
The spelling of 4432022013121 in words is "four trillion, four hundred thirty-two billion, twenty-two million, thirteen thousand, one hundred twenty-one".
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