12603143140051 has 4 divisors (see below), whose sum is σ = 12762676597600. Its totient is φ = 12443609682504.

The previous prime is 12603143140019. The next prime is 12603143140067. The reversal of 12603143140051 is 15004134130621.

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

It is a cyclic number.

It is not a de Polignac number, because 12603143140051 - 2⁵ = 12603143140019 is a prime.

It is a super-2 number, since 2×12603143140051² (a number of 27 digits) contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 12603143139995 and 12603143140022.

It is not an unprimeable number, because it can be changed into a prime (12603143144051) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 79766728656 + ... + 79766728813.

It is an arithmetic number, because the mean of its divisors is an integer number (3190669149400).

Almost surely, 2^{12603143140051} is an apocalyptic number.

12603143140051 is a deficient number, since it is larger than the sum of its proper divisors (159533457549).

12603143140051 is an equidigital number, since it uses as much as digits as its factorization.

12603143140051 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 159533457548.

The product of its (nonzero) digits is 8640, while the sum is 31.

Adding to 12603143140051 its reverse (15004134130621), we get a palindrome (27607277270672).

The spelling of 12603143140051 in words is "twelve trillion, six hundred three billion, one hundred forty-three million, one hundred forty thousand, fifty-one".