1540141304113 has 4 divisors (see below), whose sum is σ = 1551221457520. Its totient is φ = 1529061150708.

The previous prime is 1540141304089. The next prime is 1540141304129. The reversal of 1540141304113 is 3114031410451.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1540141304113 is a prime.

It is a super-3 number, since 3×1540141304113³ (a number of 38 digits) contains 333 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (1540141304153) 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, 5540076495 + ... + 5540076772.

It is an arithmetic number, because the mean of its divisors is an integer number (387805364380).

Almost surely, 2^{1540141304113} is an apocalyptic number.

It is an amenable number.

1540141304113 is a deficient number, since it is larger than the sum of its proper divisors (11080153407).

1540141304113 is a wasteful number, since it uses less digits than its factorization.

1540141304113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 11080153406.

The product of its (nonzero) digits is 2880, while the sum is 28.

Adding to 1540141304113 its reverse (3114031410451), we get a palindrome (4654172714564).

The spelling of 1540141304113 in words is "one trillion, five hundred forty billion, one hundred forty-one million, three hundred four thousand, one hundred thirteen".