13143544202550 has 96 divisors (see below), whose sum is σ = 33337709535744. Its totient is φ = 3425406192000.

The previous prime is 13143544202527. The next prime is 13143544202557. The reversal of 13143544202550 is 5520244534131.

It is a super-3 number, since 3×13143544202550³ (a number of 40 digits) contains 333 as substring.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 13143544202499 and 13143544202508.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (13143544202557) by changing a digit.

It is a polite number, since it can be written in 47 ways as a sum of consecutive naturals, for example, 3522540 + ... + 6220560.

It is an arithmetic number, because the mean of its divisors is an integer number (347267807664).

Almost surely, 2^{13143544202550} is an apocalyptic number.

13143544202550 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

It is a practical number, because each smaller number is the sum of distinct divisors of 13143544202550, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (16668854767872).

13143544202550 is an abundant number, since it is smaller than the sum of its proper divisors (20194165333194).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

13143544202550 is a wasteful number, since it uses less digits than its factorization.

13143544202550 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2698774 (or 2698769 counting only the distinct ones).

The product of its (nonzero) digits is 288000, while the sum is 39.

Adding to 13143544202550 its reverse (5520244534131), we get a palindrome (18663788736681).

The spelling of 13143544202550 in words is "thirteen trillion, one hundred forty-three billion, five hundred forty-four million, two hundred two thousand, five hundred fifty".