Base | Representation |
---|---|
bin | 1011111011111011011110… |
… | …0001001101000100110011 |
3 | 1201110122212100021200200210 |
4 | 2332332313201031010303 |
5 | 3210011332212232430 |
6 | 43525102011515203 |
7 | 2523123003365022 |
oct | 276766741150463 |
9 | 51418770250623 |
10 | 13124204024115 |
11 | 41aaa4202a408 |
12 | 157b685458b03 |
13 | 7427b7b2a002 |
14 | 3353018308b9 |
15 | 17b5cd7642b0 |
hex | befb784d133 |
13124204024115 has 16 divisors (see below), whose sum is σ = 21127552985088. Its totient is φ = 6956633297376.
The previous prime is 13124204024111. The next prime is 13124204024147. The reversal of 13124204024115 is 51142040242131.
It is not a de Polignac number, because 13124204024115 - 22 = 13124204024111 is a prime.
It is not an unprimeable number, because it can be changed into a prime (13124204024111) 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, 2683883859 + ... + 2683888748.
It is an arithmetic number, because the mean of its divisors is an integer number (1320472061568).
Almost surely, 213124204024115 is an apocalyptic number.
13124204024115 is a gapful number since it is divisible by the number (15) formed by its first and last digit.
13124204024115 is a deficient number, since it is larger than the sum of its proper divisors (8003348960973).
13124204024115 is a wasteful number, since it uses less digits than its factorization.
13124204024115 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 5367772778.
The product of its (nonzero) digits is 7680, while the sum is 30.
Adding to 13124204024115 its reverse (51142040242131), we get a palindrome (64266244266246).
The spelling of 13124204024115 in words is "thirteen trillion, one hundred twenty-four billion, two hundred four million, twenty-four thousand, one hundred fifteen".
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