Base | Representation |
---|---|
bin | 1011000001101110100111… |
… | …0100000000011111010101 |
3 | 1120221001222201220111201220 |
4 | 2300123221310000133111 |
5 | 3042121102012343013 |
6 | 41441500320240553 |
7 | 2360644651165545 |
oct | 260335164003725 |
9 | 46831881814656 |
10 | 12124320434133 |
11 | 3954993894046 |
12 | 1439936470159 |
13 | 69c41a9cc34b |
14 | 2dcb6aa4b325 |
15 | 1605ac344b23 |
hex | b06e9d007d5 |
12124320434133 has 16 divisors (see below), whose sum is σ = 16184305751040. Its totient is φ = 8073612236144.
The previous prime is 12124320434117. The next prime is 12124320434153. The reversal of 12124320434133 is 33143402342121.
It is a cyclic number.
It is not a de Polignac number, because 12124320434133 - 24 = 12124320434117 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 12124320434094 and 12124320434103.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (12124320434153) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 10173460 + ... + 11302562.
It is an arithmetic number, because the mean of its divisors is an integer number (1011519109440).
Almost surely, 212124320434133 is an apocalyptic number.
It is an amenable number.
12124320434133 is a deficient number, since it is larger than the sum of its proper divisors (4059985316907).
12124320434133 is a wasteful number, since it uses less digits than its factorization.
12124320434133 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1133208.
The product of its (nonzero) digits is 41472, while the sum is 33.
Adding to 12124320434133 its reverse (33143402342121), we get a palindrome (45267722776254).
The spelling of 12124320434133 in words is "twelve trillion, one hundred twenty-four billion, three hundred twenty million, four hundred thirty-four thousand, one hundred thirty-three".
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