Base | Representation |
---|---|
bin | 1011111011111011011110… |
… | …0001001101000101010011 |
3 | 1201110122212100021200201222 |
4 | 2332332313201031011103 |
5 | 3210011332212233042 |
6 | 43525102011515255 |
7 | 2523123003365066 |
oct | 276766741150523 |
9 | 51418770250658 |
10 | 13124204024147 |
11 | 41aaa4202a437 |
12 | 157b685458b2b |
13 | 7427b7b2a028 |
14 | 3353018308dd |
15 | 17b5cd7642d2 |
hex | befb784d153 |
13124204024147 has 2 divisors, whose sum is σ = 13124204024148. Its totient is φ = 13124204024146.
The previous prime is 13124204024111. The next prime is 13124204024189. The reversal of 13124204024147 is 74142040242131.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13124204024147 - 26 = 13124204024083 is a prime.
It is not a weakly prime, because it can be changed into another prime (13124204024947) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6562102012073 + 6562102012074.
It is an arithmetic number, because the mean of its divisors is an integer number (6562102012074).
Almost surely, 213124204024147 is an apocalyptic number.
13124204024147 is a deficient number, since it is larger than the sum of its proper divisors (1).
13124204024147 is an equidigital number, since it uses as much as digits as its factorization.
13124204024147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 43008, while the sum is 35.
Adding to 13124204024147 its reverse (74142040242131), we get a palindrome (87266244266278).
The spelling of 13124204024147 in words is "thirteen trillion, one hundred twenty-four billion, two hundred four million, twenty-four thousand, one hundred forty-seven".
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