13140221252042 has 4 divisors (see below), whose sum is σ = 19710331878066. Its totient is φ = 6570110626020.

The previous prime is 13140221252033. The next prime is 13140221252051. The reversal of 13140221252042 is 24025212204131.

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 (13140221252033) and next prime (13140221252051).

It can be written as a sum of positive squares in only one way, i.e., 7251877370761 + 5888343881281 = 2692931^2 + 2426591^2 .

It is a super-3 number, since 3×13140221252042³ (a number of 40 digits) contains 333 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13140221251996 and 13140221252014.

It is an unprimeable number.

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 as a sum of consecutive naturals, namely, 3285055313009 + ... + 3285055313012.

Almost surely, 2^{13140221252042} is an apocalyptic number.

13140221252042 is a deficient number, since it is larger than the sum of its proper divisors (6570110626024).

13140221252042 is an equidigital number, since it uses as much as digits as its factorization.

13140221252042 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6570110626023.

The product of its (nonzero) digits is 7680, while the sum is 29.

Adding to 13140221252042 its reverse (24025212204131), we get a palindrome (37165433456173).

The spelling of 13140221252042 in words is "thirteen trillion, one hundred forty billion, two hundred twenty-one million, two hundred fifty-two thousand, forty-two".