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213124433 is a prime number
BaseRepresentation
bin11001011010000…
…00010101010001
3112212000211202112
430231000111101
5414024440213
633051555105
75165346014
oct1455002521
9485024675
10213124433
11aa3375a5
125b45ba95
1335201042
142043b37b
1513a9cea8
hexcb40551

213124433 has 2 divisors, whose sum is σ = 213124434. Its totient is φ = 213124432.

The previous prime is 213124369. The next prime is 213124459. The reversal of 213124433 is 334421312.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 182466064 + 30658369 = 13508^2 + 5537^2 .

It is a cyclic number.

It is not a de Polignac number, because 213124433 - 26 = 213124369 is a prime.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 213124399 and 213124408.

It is not a weakly prime, because it can be changed into another prime (213124493) 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, 106562216 + 106562217.

It is an arithmetic number, because the mean of its divisors is an integer number (106562217).

Almost surely, 2213124433 is an apocalyptic number.

It is an amenable number.

213124433 is a deficient number, since it is larger than the sum of its proper divisors (1).

213124433 is an equidigital number, since it uses as much as digits as its factorization.

213124433 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 1728, while the sum is 23.

The square root of 213124433 is about 14598.7819012409. The cubic root of 213124433 is about 597.3255343987.

Adding to 213124433 its reverse (334421312), we get a palindrome (547545745).

The spelling of 213124433 in words is "two hundred thirteen million, one hundred twenty-four thousand, four hundred thirty-three".