Base | Representation |
---|---|
bin | 1011111100000000000111… |
… | …0100010000010100110111 |
3 | 1201110210002012210212021002 |
4 | 2333000001310100110313 |
5 | 3210021410314012101 |
6 | 43525425420513515 |
7 | 2523165620465156 |
oct | 277000164202467 |
9 | 51423065725232 |
10 | 13125450532151 |
11 | 4200521711341 |
12 | 157b9729ab89b |
13 | 742956156329 |
14 | 3353bd1c959d |
15 | 17b652dd9d6b |
hex | bf001d10537 |
13125450532151 has 2 divisors, whose sum is σ = 13125450532152. Its totient is φ = 13125450532150.
The previous prime is 13125450532147. The next prime is 13125450532153. The reversal of 13125450532151 is 15123505452131.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13125450532151 - 22 = 13125450532147 is a prime.
It is a Sophie Germain prime.
Together with 13125450532153, it forms a pair of twin primes.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13125450532153) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6562725266075 + 6562725266076.
It is an arithmetic number, because the mean of its divisors is an integer number (6562725266076).
Almost surely, 213125450532151 is an apocalyptic number.
13125450532151 is a deficient number, since it is larger than the sum of its proper divisors (1).
13125450532151 is an equidigital number, since it uses as much as digits as its factorization.
13125450532151 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 90000, while the sum is 38.
Adding to 13125450532151 its reverse (15123505452131), we get a palindrome (28248955984282).
The spelling of 13125450532151 in words is "thirteen trillion, one hundred twenty-five billion, four hundred fifty million, five hundred thirty-two thousand, one hundred fifty-one".
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