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142433523025399 is a prime number
BaseRepresentation
bin100000011000101011100011…
…100010110100010111110111
3200200022111211010010220020101
4200120223203202310113313
5122132112323343303044
61222533020203014531
742000324323513104
oct4030534342642767
9620274733126211
10142433523025399
11414247995715a1
1213b846a280a447
136162567816cb8
142725b6821baab
151170051e45cd4
hex818ae38b45f7

142433523025399 has 2 divisors, whose sum is σ = 142433523025400. Its totient is φ = 142433523025398.

The previous prime is 142433523025373. The next prime is 142433523025421. The reversal of 142433523025399 is 993520325334241.

It is a happy number.

142433523025399 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It is an emirp because it is prime and its reverse (993520325334241) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 142433523025399 - 245 = 107249150936567 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (142433523025349) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71216761512699 + 71216761512700.

It is an arithmetic number, because the mean of its divisors is an integer number (71216761512700).

Almost surely, 2142433523025399 is an apocalyptic number.

142433523025399 is a deficient number, since it is larger than the sum of its proper divisors (1).

142433523025399 is an equidigital number, since it uses as much as digits as its factorization.

142433523025399 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 20995200, while the sum is 55.

The spelling of 142433523025399 in words is "one hundred forty-two trillion, four hundred thirty-three billion, five hundred twenty-three million, twenty-five thousand, three hundred ninety-nine".