Base | Representation |
---|---|
bin | 100110100110000000110… |
… | …1101001000000000100001 |
3 | 200210002100201010021122012 |
4 | 1031030001231020000201 |
5 | 1143401213140421212 |
6 | 15140433504522305 |
7 | 1055136553266401 |
oct | 115140155100041 |
9 | 20702321107565 |
10 | 5304313217057 |
11 | 1765601589658 |
12 | 718017920395 |
13 | 2c626b68c2ca |
14 | 144a31692801 |
15 | 92e9d8c5922 |
hex | 4d301b48021 |
5304313217057 has 2 divisors, whose sum is σ = 5304313217058. Its totient is φ = 5304313217056.
The previous prime is 5304313216957. The next prime is 5304313217069. The reversal of 5304313217057 is 7507123134035.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 3970470700816 + 1333842516241 = 1992604^2 + 1154921^2 .
It is a cyclic number.
It is not a de Polignac number, because 5304313217057 - 232 = 5300018249761 is a prime.
It is not a weakly prime, because it can be changed into another prime (5304313217087) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2652156608528 + 2652156608529.
It is an arithmetic number, because the mean of its divisors is an integer number (2652156608529).
Almost surely, 25304313217057 is an apocalyptic number.
It is an amenable number.
5304313217057 is a deficient number, since it is larger than the sum of its proper divisors (1).
5304313217057 is an equidigital number, since it uses as much as digits as its factorization.
5304313217057 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 264600, while the sum is 41.
The spelling of 5304313217057 in words is "five trillion, three hundred four billion, three hundred thirteen million, two hundred seventeen thousand, fifty-seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.073 sec. • engine limits •