Base | Representation |
---|---|
bin | 1100001101011001100110… |
… | …1111000010000110101001 |
3 | 1202112100121011201212202212 |
4 | 3003112121233002012221 |
5 | 3224421033110342213 |
6 | 44315021233334505 |
7 | 2553606646430522 |
oct | 303263157020651 |
9 | 52470534655685 |
10 | 13424352043433 |
11 | 4306265835821 |
12 | 1609890407435 |
13 | 764bac5130c6 |
14 | 345a564a3849 |
15 | 1842e8e109a8 |
hex | c3599bc21a9 |
13424352043433 has 2 divisors, whose sum is σ = 13424352043434. Its totient is φ = 13424352043432.
The previous prime is 13424352043409. The next prime is 13424352043469. The reversal of 13424352043433 is 33434025342431.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 11630916830569 + 1793435212864 = 3410413^2 + 1339192^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13424352043433 is a prime.
It is not a weakly prime, because it can be changed into another prime (13424352043633) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6712176021716 + 6712176021717.
It is an arithmetic number, because the mean of its divisors is an integer number (6712176021717).
Almost surely, 213424352043433 is an apocalyptic number.
It is an amenable number.
13424352043433 is a deficient number, since it is larger than the sum of its proper divisors (1).
13424352043433 is an equidigital number, since it uses as much as digits as its factorization.
13424352043433 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1244160, while the sum is 41.
Adding to 13424352043433 its reverse (33434025342431), we get a palindrome (46858377385864).
The spelling of 13424352043433 in words is "thirteen trillion, four hundred twenty-four billion, three hundred fifty-two million, forty-three thousand, four hundred thirty-three".
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