143121001004134 has 4 divisors (see below), whose sum is σ = 214681501506204. Its totient is φ = 71560500502066.

The previous prime is 143121001004089. The next prime is 143121001004147. The reversal of 143121001004134 is 431400100121341.

It is a semiprime because it is the product of two primes.

It is a junction number, because it is equal to n+sod(n) for n = 143121001004099 and 143121001004108.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 35780250251032 + ... + 35780250251035.

It is an arithmetic number, because the mean of its divisors is an integer number (53670375376551).

Almost surely, 2^{143121001004134} is an apocalyptic number.

143121001004134 is a deficient number, since it is larger than the sum of its proper divisors (71560500502070).

143121001004134 is an equidigital number, since it uses as much as digits as its factorization.

143121001004134 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 71560500502069.

The product of its (nonzero) digits is 1152, while the sum is 25.

Adding to 143121001004134 its reverse (431400100121341), we get a palindrome (574521101125475).

The spelling of 143121001004134 in words is "one hundred forty-three trillion, one hundred twenty-one billion, one million, four thousand, one hundred thirty-four".