143431300111 has 4 divisors (see below), whose sum is σ = 156470509224. Its totient is φ = 130392091000.

The previous prime is 143431300097. The next prime is 143431300127. The reversal of 143431300111 is 111003134341.

143431300111 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 111003134341 = 11 ⋅10091194031.

It is a cyclic number.

It is not a de Polignac number, because 143431300111 - 2⁷ = 143431299983 is a prime.

It is a super-2 number, since 2×143431300111² (a number of 23 digits) contains 22 as substring.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (143431300141) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 6519604540 + ... + 6519604561.

It is an arithmetic number, because the mean of its divisors is an integer number (39117627306).

Almost surely, 2^143431300111 is an apocalyptic number.

143431300111 is a gapful number since it is divisible by the number (11) formed by its first and last digit.

143431300111 is a deficient number, since it is larger than the sum of its proper divisors (13039209113).

143431300111 is a wasteful number, since it uses less digits than its factorization.

143431300111 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 13039209112.

The product of its (nonzero) digits is 432, while the sum is 22.

Adding to 143431300111 its reverse (111003134341), we get a palindrome (254434434452).

The spelling of 143431300111 in words is "one hundred forty-three billion, four hundred thirty-one million, three hundred thousand, one hundred eleven".