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13143242111471 is a prime number
BaseRepresentation
bin1011111101000010011001…
…0001101111100111101111
31201112111000010210201212022
42333100212101233213233
53210314324440031341
643541531100100355
72524365535431623
oct277204621574757
951474003721768
1013143242111471
11420802158a1aa
1215832b92346bb
13744530141864
143361ca081183
1517bd44d23e4b
hexbf42646f9ef

13143242111471 has 2 divisors, whose sum is σ = 13143242111472. Its totient is φ = 13143242111470.

The previous prime is 13143242111423. The next prime is 13143242111533. The reversal of 13143242111471 is 17411124234131.

It is a happy number.

It is a weak prime.

It is an emirp because it is prime and its reverse (17411124234131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13143242111471 - 26 = 13143242111407 is a prime.

It is a super-3 number, since 3×131432421114713 (a number of 40 digits) contains 333 as substring.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13143242111401) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6571621055735 + 6571621055736.

It is an arithmetic number, because the mean of its divisors is an integer number (6571621055736).

Almost surely, 213143242111471 is an apocalyptic number.

13143242111471 is a deficient number, since it is larger than the sum of its proper divisors (1).

13143242111471 is an equidigital number, since it uses as much as digits as its factorization.

13143242111471 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 16128, while the sum is 35.

The spelling of 13143242111471 in words is "thirteen trillion, one hundred forty-three billion, two hundred forty-two million, one hundred eleven thousand, four hundred seventy-one".