13401210031 has 4 divisors (see below), whose sum is σ = 14619501864. Its totient is φ = 12182918200.

The previous prime is 13401210013. The next prime is 13401210049. The reversal of 13401210031 is 13001210431.

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

It is an interprime number because it is at equal distance from previous prime (13401210013) and next prime (13401210049).

It is a cyclic number.

It is not a de Polignac number, because 13401210031 - 2⁵ = 13401209999 is a prime.

It is a super-2 number, since 2×13401210031² (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13401209987 and 13401210014.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (13401210091) 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, 609145900 + ... + 609145921.

It is an arithmetic number, because the mean of its divisors is an integer number (3654875466).

Almost surely, 2^13401210031 is an apocalyptic number.

13401210031 is a gapful number since it is divisible by the number (11) formed by its first and last digit.

13401210031 is a deficient number, since it is larger than the sum of its proper divisors (1218291833).

13401210031 is a wasteful number, since it uses less digits than its factorization.

13401210031 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1218291832.

The product of its (nonzero) digits is 72, while the sum is 16.

Adding to 13401210031 its reverse (13001210431), we get a palindrome (26402420462).

The spelling of 13401210031 in words is "thirteen billion, four hundred one million, two hundred ten thousand, thirty-one".