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20124111303433 is a prime number
BaseRepresentation
bin1001001001101100000101…
…00000010111001100001001
32122020211211220202222012201
410210312002200113030021
510114203134443202213
6110444520014040201
74144630144501444
oct444660240271411
978224756688181
1020124111303433
116459651218295
1223102301a9061
13b2c907190217
144d802560c65b
1524d71ac4a6dd
hex124d82817309

20124111303433 has 2 divisors, whose sum is σ = 20124111303434. Its totient is φ = 20124111303432.

The previous prime is 20124111303421. The next prime is 20124111303493. The reversal of 20124111303433 is 33430311142102.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 12052756437264 + 8071354866169 = 3471708^2 + 2841013^2 .

It is a cyclic number.

It is not a de Polignac number, because 20124111303433 - 237 = 19986672349961 is a prime.

It is not a weakly prime, because it can be changed into another prime (20124111303403) 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, 10062055651716 + 10062055651717.

It is an arithmetic number, because the mean of its divisors is an integer number (10062055651717).

Almost surely, 220124111303433 is an apocalyptic number.

It is an amenable number.

20124111303433 is a deficient number, since it is larger than the sum of its proper divisors (1).

20124111303433 is an equidigital number, since it uses as much as digits as its factorization.

20124111303433 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 28.

Adding to 20124111303433 its reverse (33430311142102), we get a palindrome (53554422445535).

The spelling of 20124111303433 in words is "twenty trillion, one hundred twenty-four billion, one hundred eleven million, three hundred three thousand, four hundred thirty-three".