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6443350421341 is a prime number
BaseRepresentation
bin101110111000011010110…
…0111001110111101011101
3211210222102011120111102211
41131300311213032331131
51321031440201440331
621412011225120421
71233342141626221
oct135606547167535
924728364514384
106443350421341
112064678375a49
12880921029111
133797b3c07b12
14183c07502781
15b291657a8b1
hex5dc359cef5d

6443350421341 has 2 divisors, whose sum is σ = 6443350421342. Its totient is φ = 6443350421340.

The previous prime is 6443350421317. The next prime is 6443350421369. The reversal of 6443350421341 is 1431240533446.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 6337071161316 + 106279260025 = 2517354^2 + 326005^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-6443350421341 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 6443350421294 and 6443350421303.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (6443350421311) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3221675210670 + 3221675210671.

It is an arithmetic number, because the mean of its divisors is an integer number (3221675210671).

Almost surely, 26443350421341 is an apocalyptic number.

It is an amenable number.

6443350421341 is a deficient number, since it is larger than the sum of its proper divisors (1).

6443350421341 is an equidigital number, since it uses as much as digits as its factorization.

6443350421341 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 414720, while the sum is 40.

Adding to 6443350421341 its reverse (1431240533446), we get a palindrome (7874590954787).

The spelling of 6443350421341 in words is "six trillion, four hundred forty-three billion, three hundred fifty million, four hundred twenty-one thousand, three hundred forty-one".