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435130274081 is a prime number
BaseRepresentation
bin1100101010011111100…
…10010001000100100001
31112121010222101012101002
412111033302101010201
524112121322232311
6531521315521345
743302635206106
oct6251762210441
91477128335332
10435130274081
111585a0772268
12703b82b9255
1332056845479
14170bbb00aad
15b4bab0d93b
hex654fc91121

435130274081 has 2 divisors, whose sum is σ = 435130274082. Its totient is φ = 435130274080.

The previous prime is 435130274077. The next prime is 435130274083. The reversal of 435130274081 is 180472031534.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 333640243456 + 101490030625 = 577616^2 + 318575^2 .

It is a cyclic number.

It is not a de Polignac number, because 435130274081 - 22 = 435130274077 is a prime.

It is a super-2 number, since 2×4351302740812 (a number of 24 digits) contains 22 as substring.

It is a Sophie Germain prime.

Together with 435130274083, it forms a pair of twin primes.

It is a Chen prime.

It is a Curzon number.

It is not a weakly prime, because it can be changed into another prime (435130274083) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 217565137040 + 217565137041.

It is an arithmetic number, because the mean of its divisors is an integer number (217565137041).

Almost surely, 2435130274081 is an apocalyptic number.

It is an amenable number.

435130274081 is a deficient number, since it is larger than the sum of its proper divisors (1).

435130274081 is an equidigital number, since it uses as much as digits as its factorization.

435130274081 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 80640, while the sum is 38.

The spelling of 435130274081 in words is "four hundred thirty-five billion, one hundred thirty million, two hundred seventy-four thousand, eighty-one".