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534012433 is a prime number
BaseRepresentation
bin11111110101000…
…110001000010001
31101012211122100021
4133311012020101
52043201344213
6124553423441
716143013363
oct3765061021
91335748307
10534012433
11254488499
1212aa0ab81
13868334b6
1450ccad33
1531d35d8d
hex1fd46211

534012433 has 2 divisors, whose sum is σ = 534012434. Its totient is φ = 534012432.

The previous prime is 534012407. The next prime is 534012439. The reversal of 534012433 is 334210435.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 521985409 + 12027024 = 22847^2 + 3468^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-534012433 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 534012398 and 534012407.

It is not a weakly prime, because it can be changed into another prime (534012439) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 267006216 + 267006217.

It is an arithmetic number, because the mean of its divisors is an integer number (267006217).

Almost surely, 2534012433 is an apocalyptic number.

It is an amenable number.

534012433 is a deficient number, since it is larger than the sum of its proper divisors (1).

534012433 is an equidigital number, since it uses as much as digits as its factorization.

534012433 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 4320, while the sum is 25.

The square root of 534012433 is about 23108.7090292816. The cubic root of 534012433 is about 811.3043218666.

Adding to 534012433 its reverse (334210435), we get a palindrome (868222868).

The spelling of 534012433 in words is "five hundred thirty-four million, twelve thousand, four hundred thirty-three".