Base | Representation |
---|---|
bin | 10010111111011011001… |
… | …110101111010001101011 |
3 | 11121202120101202210222111 |
4 | 102333123032233101223 |
5 | 132340222102232101 |
6 | 2435311054303151 |
7 | 163200301604005 |
oct | 22773316572153 |
9 | 4552511683874 |
10 | 1305053164651 |
11 | 463519541625 |
12 | 190b177b7ab7 |
13 | 960b1c7b9b2 |
14 | 47244729375 |
15 | 23e3287e251 |
hex | 12fdb3af46b |
1305053164651 has 2 divisors, whose sum is σ = 1305053164652. Its totient is φ = 1305053164650.
The previous prime is 1305053164633. The next prime is 1305053164663. The reversal of 1305053164651 is 1564613505031.
It is a strong prime.
It is an emirp because it is prime and its reverse (1564613505031) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1305053164651 - 25 = 1305053164619 is a prime.
It is not a weakly prime, because it can be changed into another prime (1305053164691) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 652526582325 + 652526582326.
It is an arithmetic number, because the mean of its divisors is an integer number (652526582326).
Almost surely, 21305053164651 is an apocalyptic number.
1305053164651 is a deficient number, since it is larger than the sum of its proper divisors (1).
1305053164651 is an equidigital number, since it uses as much as digits as its factorization.
1305053164651 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 162000, while the sum is 40.
Adding to 1305053164651 its reverse (1564613505031), we get a palindrome (2869666669682).
The spelling of 1305053164651 in words is "one trillion, three hundred five billion, fifty-three million, one hundred sixty-four thousand, six hundred fifty-one".
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