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13300242433 is a prime number
BaseRepresentation
bin11000110001100000…
…11001100000000001
31021022220202022122121
430120300121200001
5204214330224213
610035434204241
7650410124002
oct143060314001
937286668577
1013300242433
115705704aa5
1226b2283681
13133c655c4a
149025a31a9
1552c9ac88d
hex318c19801

13300242433 has 2 divisors, whose sum is σ = 13300242434. Its totient is φ = 13300242432.

The previous prime is 13300242419. The next prime is 13300242533. The reversal of 13300242433 is 33424200331.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11368464129 + 1931778304 = 106623^2 + 43952^2 .

It is a cyclic number.

It is not a de Polignac number, because 13300242433 - 221 = 13298145281 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 13300242398 and 13300242407.

It is not a weakly prime, because it can be changed into another prime (13300242413) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (11) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6650121216 + 6650121217.

It is an arithmetic number, because the mean of its divisors is an integer number (6650121217).

Almost surely, 213300242433 is an apocalyptic number.

It is an amenable number.

13300242433 is a deficient number, since it is larger than the sum of its proper divisors (1).

13300242433 is an equidigital number, since it uses as much as digits as its factorization.

13300242433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 5184, while the sum is 25.

Adding to 13300242433 its reverse (33424200331), we get a palindrome (46724442764).

The spelling of 13300242433 in words is "thirteen billion, three hundred million, two hundred forty-two thousand, four hundred thirty-three".