Base | Representation |
---|---|
bin | 11000011100100010… |
… | …11011010010110001 |
3 | 1020212122210010011202 |
4 | 30032101123102301 |
5 | 203334313214103 |
6 | 10010152220545 |
7 | 643143100304 |
oct | 141621332261 |
9 | 36778703152 |
10 | 13124351153 |
11 | 5625399301 |
12 | 266339a755 |
13 | 1312090274 |
14 | 8c7098b3b |
15 | 51c316988 |
hex | 30e45b4b1 |
13124351153 has 2 divisors, whose sum is σ = 13124351154. Its totient is φ = 13124351152.
The previous prime is 13124351077. The next prime is 13124351159. The reversal of 13124351153 is 35115342131.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 13054662049 + 69689104 = 114257^2 + 8348^2 .
It is an emirp because it is prime and its reverse (35115342131) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13124351153 is a prime.
It is not a weakly prime, because it can be changed into another prime (13124351159) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6562175576 + 6562175577.
It is an arithmetic number, because the mean of its divisors is an integer number (6562175577).
Almost surely, 213124351153 is an apocalyptic number.
It is an amenable number.
13124351153 is a deficient number, since it is larger than the sum of its proper divisors (1).
13124351153 is an equidigital number, since it uses as much as digits as its factorization.
13124351153 is an evil number, because the sum of its binary digits is even.
The product of its digits is 5400, while the sum is 29.
Adding to 13124351153 its reverse (35115342131), we get a palindrome (48239693284).
The spelling of 13124351153 in words is "thirteen billion, one hundred twenty-four million, three hundred fifty-one thousand, one hundred fifty-three".
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