Base | Representation |
---|---|
bin | 100110111000001110101… |
… | …1100011010110000001001 |
3 | 200220211100001110122211211 |
4 | 1031300131130122300021 |
5 | 1200021322411230342 |
6 | 15210423412053121 |
7 | 1061023154455246 |
oct | 115603534326011 |
9 | 20824301418754 |
10 | 5343433305097 |
11 | 1780156885828 |
12 | 7237150041a1 |
13 | 2c9b6435c6b4 |
14 | 1468a30525cd |
15 | 93edceecc17 |
hex | 4dc1d71ac09 |
5343433305097 has 2 divisors, whose sum is σ = 5343433305098. Its totient is φ = 5343433305096.
The previous prime is 5343433305037. The next prime is 5343433305127. The reversal of 5343433305097 is 7905033343435.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4635284126841 + 708149178256 = 2152971^2 + 841516^2 .
It is a cyclic number.
It is not a de Polignac number, because 5343433305097 - 211 = 5343433303049 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 5343433305097.
It is not a weakly prime, because it can be changed into another prime (5343433305037) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2671716652548 + 2671716652549.
It is an arithmetic number, because the mean of its divisors is an integer number (2671716652549).
Almost surely, 25343433305097 is an apocalyptic number.
It is an amenable number.
5343433305097 is a deficient number, since it is larger than the sum of its proper divisors (1).
5343433305097 is an equidigital number, since it uses as much as digits as its factorization.
5343433305097 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6123600, while the sum is 49.
The spelling of 5343433305097 in words is "five trillion, three hundred forty-three billion, four hundred thirty-three million, three hundred five thousand, ninety-seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.079 sec. • engine limits •