Base | Representation |
---|---|
bin | 11011110001010100… |
… | …00111110101101011 |
3 | 1102111001022100011121 |
4 | 31320222013311223 |
5 | 221013224204323 |
6 | 10503232040111 |
7 | 1035315123304 |
oct | 157052076553 |
9 | 42431270147 |
10 | 14909209963 |
11 | 636095112a |
12 | 2a81093037 |
13 | 1537aab63b |
14 | a1614d7ab |
15 | 5c3d7da5d |
hex | 378a87d6b |
14909209963 has 2 divisors, whose sum is σ = 14909209964. Its totient is φ = 14909209962.
The previous prime is 14909209961. The next prime is 14909209979. The reversal of 14909209963 is 36990290941.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 14909209963 - 21 = 14909209961 is a prime.
It is a super-2 number, since 2×149092099632 (a number of 21 digits) contains 22 as substring.
Together with 14909209961, it forms a pair of twin primes.
It is a self number, because there is not a number n which added to its sum of digits gives 14909209963.
It is not a weakly prime, because it can be changed into another prime (14909209961) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7454604981 + 7454604982.
It is an arithmetic number, because the mean of its divisors is an integer number (7454604982).
Almost surely, 214909209963 is an apocalyptic number.
14909209963 is a deficient number, since it is larger than the sum of its proper divisors (1).
14909209963 is an equidigital number, since it uses as much as digits as its factorization.
14909209963 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 944784, while the sum is 52.
The spelling of 14909209963 in words is "fourteen billion, nine hundred nine million, two hundred nine thousand, nine hundred sixty-three".
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