Base | Representation |
---|---|
bin | 100101010010000110101… |
… | …1110101110010101000001 |
3 | 200010212020212211020201012 |
4 | 1022210031132232111001 |
5 | 1132423200232021032 |
6 | 14521553400450305 |
7 | 1036130344363244 |
oct | 112441536562501 |
9 | 20125225736635 |
10 | 5124122142017 |
11 | 16a61453aaa1a |
12 | 6a910a179995 |
13 | 2b2283296c08 |
14 | 13a01a35925b |
15 | 8d4545276b2 |
hex | 4a90d7ae541 |
5124122142017 has 2 divisors, whose sum is σ = 5124122142018. Its totient is φ = 5124122142016.
The previous prime is 5124122141951. The next prime is 5124122142019. The reversal of 5124122142017 is 7102412214215.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4290847102096 + 833275039921 = 2071436^2 + 912839^2 .
It is a cyclic number.
It is not a de Polignac number, because 5124122142017 - 216 = 5124122076481 is a prime.
Together with 5124122142019, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (5124122142019) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2562061071008 + 2562061071009.
It is an arithmetic number, because the mean of its divisors is an integer number (2562061071009).
Almost surely, 25124122142017 is an apocalyptic number.
It is an amenable number.
5124122142017 is a deficient number, since it is larger than the sum of its proper divisors (1).
5124122142017 is an equidigital number, since it uses as much as digits as its factorization.
5124122142017 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 8960, while the sum is 32.
The spelling of 5124122142017 in words is "five trillion, one hundred twenty-four billion, one hundred twenty-two million, one hundred forty-two thousand, seventeen".
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