Base | Representation |
---|---|
bin | 1100001001010001011101… |
… | …0000001101111010110111 |
3 | 1202021120120102201010210001 |
4 | 3002110113100031322313 |
5 | 3222240323010404201 |
6 | 44222253124212131 |
7 | 2545516530353161 |
oct | 302242720157267 |
9 | 52246512633701 |
10 | 13353443450551 |
11 | 4289188782446 |
12 | 15b7ba123a047 |
13 | 75b2bba83b39 |
14 | 34244ad35931 |
15 | 182548b77301 |
hex | c251740deb7 |
13353443450551 has 2 divisors, whose sum is σ = 13353443450552. Its totient is φ = 13353443450550.
The previous prime is 13353443450497. The next prime is 13353443450603. The reversal of 13353443450551 is 15505434435331.
13353443450551 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13353443450551 is a prime.
It is a super-3 number, since 3×133534434505513 (a number of 40 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13353443450051) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6676721725275 + 6676721725276.
It is an arithmetic number, because the mean of its divisors is an integer number (6676721725276).
Almost surely, 213353443450551 is an apocalyptic number.
13353443450551 is a deficient number, since it is larger than the sum of its proper divisors (1).
13353443450551 is an equidigital number, since it uses as much as digits as its factorization.
13353443450551 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 3240000, while the sum is 46.
Adding to 13353443450551 its reverse (15505434435331), we get a palindrome (28858877885882).
The spelling of 13353443450551 in words is "thirteen trillion, three hundred fifty-three billion, four hundred forty-three million, four hundred fifty thousand, five hundred fifty-one".
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