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12143023411319 is a prime number
BaseRepresentation
bin1011000010110100010010…
…0110001101000001110111
31120222212021011212012011122
42300231010212031001313
53042422402443130234
641454232234054155
72362206303652043
oct260550446150167
946885234765148
1012143023411319
113961911169689
121441495b7a35b
136a110b772158
142dda22989023
15160d042ac92e
hexb0b4498d077

12143023411319 has 2 divisors, whose sum is σ = 12143023411320. Its totient is φ = 12143023411318.

The previous prime is 12143023411301. The next prime is 12143023411357. The reversal of 12143023411319 is 91311432034121.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 12143023411319 - 232 = 12138728444023 is a prime.

It is a super-3 number, since 3×121430234113193 (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (12143023411399) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6071511705659 + 6071511705660.

It is an arithmetic number, because the mean of its divisors is an integer number (6071511705660).

Almost surely, 212143023411319 is an apocalyptic number.

12143023411319 is a deficient number, since it is larger than the sum of its proper divisors (1).

12143023411319 is an equidigital number, since it uses as much as digits as its factorization.

12143023411319 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 15552, while the sum is 35.

The spelling of 12143023411319 in words is "twelve trillion, one hundred forty-three billion, twenty-three million, four hundred eleven thousand, three hundred nineteen".