Base | Representation |
---|---|
bin | 10110010101001111010111… |
… | …00101010101011010011011 |
3 | 20102212201010210202011022121 |
4 | 23022213223211111122123 |
5 | 22414042311230343323 |
6 | 252240014153402111 |
7 | 13225646534442334 |
oct | 1312475345253233 |
9 | 212781123664277 |
10 | 49108313527963 |
11 | 1471379a80a968 |
12 | 561162aa09337 |
13 | 2152b814333b1 |
14 | c1abd83c128b |
15 | 5a2647410a5d |
hex | 2ca9eb95569b |
49108313527963 has 2 divisors, whose sum is σ = 49108313527964. Its totient is φ = 49108313527962.
The previous prime is 49108313527957. The next prime is 49108313528003. The reversal of 49108313527963 is 36972531380194.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-49108313527963 is a prime.
It is not a weakly prime, because it can be changed into another prime (49108313522963) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 24554156763981 + 24554156763982.
It is an arithmetic number, because the mean of its divisors is an integer number (24554156763982).
It is a 1-persistent number, because it is pandigital, but 2⋅49108313527963 = 98216627055926 is not.
Almost surely, 249108313527963 is an apocalyptic number.
49108313527963 is a deficient number, since it is larger than the sum of its proper divisors (1).
49108313527963 is an equidigital number, since it uses as much as digits as its factorization.
49108313527963 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 29393280, while the sum is 61.
The spelling of 49108313527963 in words is "forty-nine trillion, one hundred eight billion, three hundred thirteen million, five hundred twenty-seven thousand, nine hundred sixty-three".
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