Base | Representation |
---|---|
bin | 1100001001001110001000… |
… | …0100010101001011111111 |
3 | 1202021111021020200011202111 |
4 | 3002103202010111023333 |
5 | 3222232011043022033 |
6 | 44222024411034451 |
7 | 2545455444021013 |
oct | 302234204251377 |
9 | 52244236604674 |
10 | 13352551142143 |
11 | 4288870023984 |
12 | 15b799243ba27 |
13 | 75b1a8c513b5 |
14 | 3423a461c543 |
15 | 1824e56697cd |
hex | c24e21152ff |
13352551142143 has 2 divisors, whose sum is σ = 13352551142144. Its totient is φ = 13352551142142.
The previous prime is 13352551142071. The next prime is 13352551142149. The reversal of 13352551142143 is 34124115525331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13352551142143 - 237 = 13215112188671 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13352551142096 and 13352551142105.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13352551142149) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6676275571071 + 6676275571072.
It is an arithmetic number, because the mean of its divisors is an integer number (6676275571072).
Almost surely, 213352551142143 is an apocalyptic number.
13352551142143 is a deficient number, since it is larger than the sum of its proper divisors (1).
13352551142143 is an equidigital number, since it uses as much as digits as its factorization.
13352551142143 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 216000, while the sum is 40.
Adding to 13352551142143 its reverse (34124115525331), we get a palindrome (47476666667474).
The spelling of 13352551142143 in words is "thirteen trillion, three hundred fifty-two billion, five hundred fifty-one million, one hundred forty-two thousand, one hundred forty-three".
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