Base | Representation |
---|---|
bin | 110100100110110001… |
… | …111100010100110111 |
3 | 12101210112212112020011 |
4 | 310212301330110313 |
5 | 1411140202414312 |
6 | 41540545521051 |
7 | 4036520200621 |
oct | 644661742467 |
9 | 171715775204 |
10 | 56485201207 |
11 | 21a56472776 |
12 | ab44943187 |
13 | 543250b9a6 |
14 | 2a3bb61c11 |
15 | 1708dc0da7 |
hex | d26c7c537 |
56485201207 has 2 divisors, whose sum is σ = 56485201208. Its totient is φ = 56485201206.
The previous prime is 56485201159. The next prime is 56485201217. The reversal of 56485201207 is 70210258465.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-56485201207 is a prime.
It is a super-3 number, since 3×564852012073 (a number of 33 digits) contains 333 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 56485201207.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (56485201217) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 28242600603 + 28242600604.
It is an arithmetic number, because the mean of its divisors is an integer number (28242600604).
Almost surely, 256485201207 is an apocalyptic number.
56485201207 is a deficient number, since it is larger than the sum of its proper divisors (1).
56485201207 is an equidigital number, since it uses as much as digits as its factorization.
56485201207 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 134400, while the sum is 40.
The spelling of 56485201207 in words is "fifty-six billion, four hundred eighty-five million, two hundred one thousand, two hundred seven".
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