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530010113 is a prime number
BaseRepresentation
bin11111100101110…
…101000000000001
31100221022022012012
4133211311000001
52041140310423
6124331542305
716064003663
oct3745650001
91327268165
10530010113
112521a4496
121295ba995
1385a61869
1450568533
15317e5078
hex1f975001

530010113 has 2 divisors, whose sum is σ = 530010114. Its totient is φ = 530010112.

The previous prime is 530010097. The next prime is 530010161. The reversal of 530010113 is 311010035.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 397483969 + 132526144 = 19937^2 + 11512^2 .

It is a cyclic number.

It is not a de Polignac number, because 530010113 - 24 = 530010097 is a prime.

It is not a weakly prime, because it can be changed into another prime (530010713) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265005056 + 265005057.

It is an arithmetic number, because the mean of its divisors is an integer number (265005057).

Almost surely, 2530010113 is an apocalyptic number.

It is an amenable number.

530010113 is a deficient number, since it is larger than the sum of its proper divisors (1).

530010113 is an equidigital number, since it uses as much as digits as its factorization.

530010113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 45, while the sum is 14.

The square root of 530010113 is about 23021.9485057195. The cubic root of 530010113 is about 809.2723806689.

Adding to 530010113 its reverse (311010035), we get a palindrome (841020148).

The spelling of 530010113 in words is "five hundred thirty million, ten thousand, one hundred thirteen".