11953543643 has 2 divisors, whose sum is σ = 11953543644. Its totient is φ = 11953543642.

The previous prime is 11953543619. The next prime is 11953543687. The reversal of 11953543643 is 34634535911.

It is an a-pointer prime, because the next prime (11953543687) can be obtained adding 11953543643 to its sum of digits (44).

It is a weak prime.

It is an emirp because it is prime and its reverse (34634535911) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 11953543643 - 2⁶ = 11953543579 is a prime.

It is a super-2 number, since 2×11953543643² (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 11953543594 and 11953543603.

It is not a weakly prime, because it can be changed into another prime (11953543243) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5976771821 + 5976771822.

It is an arithmetic number, because the mean of its divisors is an integer number (5976771822).

Almost surely, 2^11953543643 is an apocalyptic number.

11953543643 is a deficient number, since it is larger than the sum of its proper divisors (1).

11953543643 is an equidigital number, since it uses as much as digits as its factorization.

11953543643 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 583200, while the sum is 44.

The spelling of 11953543643 in words is "eleven billion, nine hundred fifty-three million, five hundred forty-three thousand, six hundred forty-three".