Base | Representation |
---|---|
bin | 100111000001110011101… |
… | …0110101010001010001001 |
3 | 200222210102102120122122212 |
4 | 1032003213112222022021 |
5 | 1200340433132142103 |
6 | 15224104325554505 |
7 | 1062351636224330 |
oct | 116034726521211 |
9 | 20883372518585 |
10 | 5364000662153 |
11 | 1788950597581 |
12 | 7276b4b75a35 |
13 | 2cba91443c63 |
14 | 147894824d17 |
15 | 947e39897d8 |
hex | 4e0e75aa289 |
5364000662153 has 16 divisors (see below), whose sum is σ = 6296065062144. Its totient is φ = 4473384667200.
The previous prime is 5364000662149. The next prime is 5364000662197. The reversal of 5364000662153 is 3512660004635.
It is a cyclic number.
It is not a de Polignac number, because 5364000662153 - 22 = 5364000662149 is a prime.
It is a Duffinian number.
It is a self number, because there is not a number n which added to its sum of digits gives 5364000662153.
It is not an unprimeable number, because it can be changed into a prime (5364000662123) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 24483251 + ... + 24701367.
It is an arithmetic number, because the mean of its divisors is an integer number (393504066384).
Almost surely, 25364000662153 is an apocalyptic number.
It is an amenable number.
5364000662153 is a deficient number, since it is larger than the sum of its proper divisors (932064399991).
5364000662153 is a wasteful number, since it uses less digits than its factorization.
5364000662153 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 313112.
The product of its (nonzero) digits is 388800, while the sum is 41.
Adding to 5364000662153 its reverse (3512660004635), we get a palindrome (8876660666788).
The spelling of 5364000662153 in words is "five trillion, three hundred sixty-four billion, six hundred sixty-two thousand, one hundred fifty-three".
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