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113021424433 is a prime number
BaseRepresentation
bin110100101000010011…
…0100001001100110001
3101210201122202020111201
41221100212201030301
53322431441040213
6123531001211201
711110530154021
oct1512046411461
9353648666451
10113021424433
1143a28713aa5
1219aa2790b01
13a872473008
1456825a5a81
152e174be6dd
hex1a509a1331

113021424433 has 2 divisors, whose sum is σ = 113021424434. Its totient is φ = 113021424432.

The previous prime is 113021424419. The next prime is 113021424463. The reversal of 113021424433 is 334424120311.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 96915783969 + 16105640464 = 311313^2 + 126908^2 .

It is a cyclic number.

It is not a de Polignac number, because 113021424433 - 217 = 113021293361 is a prime.

It is not a weakly prime, because it can be changed into another prime (113021424463) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 56510712216 + 56510712217.

It is an arithmetic number, because the mean of its divisors is an integer number (56510712217).

Almost surely, 2113021424433 is an apocalyptic number.

It is an amenable number.

113021424433 is a deficient number, since it is larger than the sum of its proper divisors (1).

113021424433 is an equidigital number, since it uses as much as digits as its factorization.

113021424433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 6912, while the sum is 28.

Adding to 113021424433 its reverse (334424120311), we get a palindrome (447445544744).

The spelling of 113021424433 in words is "one hundred thirteen billion, twenty-one million, four hundred twenty-four thousand, four hundred thirty-three".