Base | Representation |
---|---|
bin | 1100000000100110000000… |
… | …1011001110011000101011 |
3 | 1201202022201100221122112001 |
4 | 3000021200023032120223 |
5 | 3212314443303220042 |
6 | 44025554052431431 |
7 | 2531660633550433 |
oct | 300114013163053 |
9 | 51668640848461 |
10 | 13204343023147 |
11 | 4230a26444521 |
12 | 159310b918577 |
13 | 74a2199ab673 |
14 | 339144c2d2c3 |
15 | 17d71e03cbb7 |
hex | c02602ce62b |
13204343023147 has 2 divisors, whose sum is σ = 13204343023148. Its totient is φ = 13204343023146.
The previous prime is 13204343023093. The next prime is 13204343023159. The reversal of 13204343023147 is 74132034340231.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13204343023147 is a prime.
It is not a weakly prime, because it can be changed into another prime (13204343023747) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6602171511573 + 6602171511574.
It is an arithmetic number, because the mean of its divisors is an integer number (6602171511574).
Almost surely, 213204343023147 is an apocalyptic number.
13204343023147 is a deficient number, since it is larger than the sum of its proper divisors (1).
13204343023147 is an equidigital number, since it uses as much as digits as its factorization.
13204343023147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 145152, while the sum is 37.
Adding to 13204343023147 its reverse (74132034340231), we get a palindrome (87336377363378).
The spelling of 13204343023147 in words is "thirteen trillion, two hundred four billion, three hundred forty-three million, twenty-three thousand, one hundred forty-seven".
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