40162521512031 has 4 divisors (see below), whose sum is σ = 53550028682712. Its totient is φ = 26775014341352.

The previous prime is 40162521511993. The next prime is 40162521512081. The reversal of 40162521512031 is 13021512526104.

40162521512031 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.

It is a cyclic number.

It is not a de Polignac number, because 40162521512031 - 2³⁰ = 40161447770207 is a prime.

It is a super-3 number, since 3×40162521512031³ (a number of 42 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 40162521511983 and 40162521512001.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (40162521512081) by changing a digit.

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 in 3 ways as a sum of consecutive naturals, for example, 6693753585336 + ... + 6693753585341.

It is an arithmetic number, because the mean of its divisors is an integer number (13387507170678).

Almost surely, 2^{40162521512031} is an apocalyptic number.

40162521512031 is a deficient number, since it is larger than the sum of its proper divisors (13387507170681).

40162521512031 is a wasteful number, since it uses less digits than its factorization.

40162521512031 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 13387507170680.

The product of its (nonzero) digits is 14400, while the sum is 33.

The spelling of 40162521512031 in words is "forty trillion, one hundred sixty-two billion, five hundred twenty-one million, five hundred twelve thousand, thirty-one".