100433041341041 has 2 divisors, whose sum is σ = 100433041341042. Its totient is φ = 100433041341040.

The previous prime is 100433041340971. The next prime is 100433041341043. The reversal of 100433041341041 is 140143140334001.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 96673553017441 + 3759488323600 = 9832271^2 + 1938940^2 .

It is a cyclic number.

It is not a de Polignac number, because 100433041341041 - 2¹⁸ = 100433041078897 is a prime.

It is a super-3 number, since 3×100433041341041³ (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

Together with 100433041341043, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 100433041340995 and 100433041341013.

It is not a weakly prime, because it can be changed into another prime (100433041341043) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50216520670520 + 50216520670521.

It is an arithmetic number, because the mean of its divisors is an integer number (50216520670521).

Almost surely, 2^{100433041341041} is an apocalyptic number.

It is an amenable number.

100433041341041 is a deficient number, since it is larger than the sum of its proper divisors (1).

100433041341041 is an equidigital number, since it uses as much as digits as its factorization.

100433041341041 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 6912, while the sum is 29.

Adding to 100433041341041 its reverse (140143140334001), we get a palindrome (240576181675042).

The spelling of 100433041341041 in words is "one hundred trillion, four hundred thirty-three billion, forty-one million, three hundred forty-one thousand, forty-one".