13450433405431 has 4 divisors (see below), whose sum is σ = 14673200078664. Its totient is φ = 12227666732200.

The previous prime is 13450433405429. The next prime is 13450433405443.

13450433405431 is nontrivially palindromic in base 10.

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 13450433405431 - 2¹ = 13450433405429 is a prime.

It is a super-2 number, since 2×13450433405431² (a number of 27 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 (13450433405411) 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, 611383336600 + ... + 611383336621.

It is an arithmetic number, because the mean of its divisors is an integer number (3668300019666).

Almost surely, 2^{13450433405431} is an apocalyptic number.

13450433405431 is a gapful number since it is divisible by the number (11) formed by its first and last digit.

13450433405431 is a deficient number, since it is larger than the sum of its proper divisors (1222766673233).

13450433405431 is a wasteful number, since it uses less digits than its factorization.

13450433405431 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1222766673232.

The product of its (nonzero) digits is 518400, while the sum is 40.

The spelling of 13450433405431 in words is "thirteen trillion, four hundred fifty billion, four hundred thirty-three million, four hundred five thousand, four hundred thirty-one".