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5322433124117 is a prime number
BaseRepresentation
bin100110101110011100110…
…1111000101011100010101
3200211211010111220221020002
41031130321233011130111
51144200320334432432
615153031513221045
71056350566044644
oct115347157053425
920754114827202
105322433124117
11177225a819a5a
1271b634101785
132c7b986b33c5
1414586ddb335b
15936ae566e62
hex4d739bc5715

5322433124117 has 2 divisors, whose sum is σ = 5322433124118. Its totient is φ = 5322433124116.

The previous prime is 5322433124083. The next prime is 5322433124179. The reversal of 5322433124117 is 7114213342235.

5322433124117 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 4466569457476 + 855863666641 = 2113426^2 + 925129^2 .

It is a cyclic number.

It is not a de Polignac number, because 5322433124117 - 242 = 924386613013 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (5322433124417) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2661216562058 + 2661216562059.

It is an arithmetic number, because the mean of its divisors is an integer number (2661216562059).

Almost surely, 25322433124117 is an apocalyptic number.

It is an amenable number.

5322433124117 is a deficient number, since it is larger than the sum of its proper divisors (1).

5322433124117 is an equidigital number, since it uses as much as digits as its factorization.

5322433124117 is an evil number, because the sum of its binary digits is even.

The product of its digits is 120960, while the sum is 38.

The spelling of 5322433124117 in words is "five trillion, three hundred twenty-two billion, four hundred thirty-three million, one hundred twenty-four thousand, one hundred seventeen".