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131343444024599 is a prime number
BaseRepresentation
bin11101110111010011000111…
…010100110001110100010111
3122020001022101102221212002002
4131313103013110301310113
5114203412333132241344
61143202202321505515
736444151661411042
oct3567230724616427
9566038342855062
10131343444024599
1138939495043084
12128932a57a229b
135839831100351
1424610b14a1c59
15102b82750744e
hex7774c7531d17

131343444024599 has 2 divisors, whose sum is σ = 131343444024600. Its totient is φ = 131343444024598.

The previous prime is 131343444024589. The next prime is 131343444024601. The reversal of 131343444024599 is 995420444343131.

It is a strong prime.

It is an emirp because it is prime and its reverse (995420444343131) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-131343444024599 is a prime.

It is a super-2 number, since 2×1313434440245992 (a number of 29 digits) contains 22 as substring.

Together with 131343444024601, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (131343444024539) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65671722012299 + 65671722012300.

It is an arithmetic number, because the mean of its divisors is an integer number (65671722012300).

Almost surely, 2131343444024599 is an apocalyptic number.

131343444024599 is a deficient number, since it is larger than the sum of its proper divisors (1).

131343444024599 is an equidigital number, since it uses as much as digits as its factorization.

131343444024599 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 22394880, while the sum is 56.

The spelling of 131343444024599 in words is "one hundred thirty-one trillion, three hundred forty-three billion, four hundred forty-four million, twenty-four thousand, five hundred ninety-nine".