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433131323124047 is a prime number
BaseRepresentation
bin110001001111011100011111…
…0101001111110100101001111
32002210120221122121212001200012
41202132320332221332211033
5423232410422204432142
64133105441040023435
7160142463650613542
oct14236707651764517
92083527577761605
10433131323124047
11116010a652141a9
12406b3903901b7b
131578b154534a52
1478d594c9b6859
153511b1671b882
hex189ee3ea7e94f

433131323124047 has 2 divisors, whose sum is σ = 433131323124048. Its totient is φ = 433131323124046.

The previous prime is 433131323124029. The next prime is 433131323124079. The reversal of 433131323124047 is 740421323131334.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 433131323124047 - 214 = 433131323107663 is a prime.

It is a super-2 number, since 2×4331313231240472 (a number of 30 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 433131323123995 and 433131323124013.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (433131323124347) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 216565661562023 + 216565661562024.

It is an arithmetic number, because the mean of its divisors is an integer number (216565661562024).

Almost surely, 2433131323124047 is an apocalyptic number.

433131323124047 is a deficient number, since it is larger than the sum of its proper divisors (1).

433131323124047 is an equidigital number, since it uses as much as digits as its factorization.

433131323124047 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 435456, while the sum is 41.

The spelling of 433131323124047 in words is "four hundred thirty-three trillion, one hundred thirty-one billion, three hundred twenty-three million, one hundred twenty-four thousand, forty-seven".