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BaseRepresentation
bin1001100011100011…
…00001101110100001
3111020112001201210021
410301301201232201
541001242042311
62205014233441
7241061366404
oct46161415641
914215051707
105130034081
1121a2852597
12bb2055881
13639a89a33
1436946dd3b
15200590171
hex131c61ba1

5130034081 has 2 divisors, whose sum is σ = 5130034082. Its totient is φ = 5130034080.

The previous prime is 5130034033. The next prime is 5130034111. The reversal of 5130034081 is 1804300315.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 3533232481 + 1596801600 = 59441^2 + 39960^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-5130034081 is a prime.

It is a super-2 number, since 2×51300340812 = 52634499344443029122, which contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (5130334081) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2565017040 + 2565017041.

It is an arithmetic number, because the mean of its divisors is an integer number (2565017041).

Almost surely, 25130034081 is an apocalyptic number.

It is an amenable number.

5130034081 is a deficient number, since it is larger than the sum of its proper divisors (1).

5130034081 is an equidigital number, since it uses as much as digits as its factorization.

5130034081 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1440, while the sum is 25.

The square root of 5130034081 is about 71624.2562334856. The cubic root of 5130034081 is about 1724.6729429374.

Adding to 5130034081 its reverse (1804300315), we get a palindrome (6934334396).

The spelling of 5130034081 in words is "five billion, one hundred thirty million, thirty-four thousand, eighty-one".