141132140000433 has 6 divisors (see below), whose sum is σ = 203857535556194. Its totient is φ = 94088093333616.

The previous prime is 141132140000429. The next prime is 141132140000587. The reversal of 141132140000433 is 334000041231141.

141132140000433 is a `hidden beast` number, since 1 + 4 + 1 + 13 + 214 + 0 + 0 + 0 + 0 + 433 = 666.

It can be written as a sum of positive squares in only one way, i.e., 129493862334369 + 11638277666064 = 11379537^2 + 3411492^2 .

It is not a de Polignac number, because 141132140000433 - 2² = 141132140000429 is a prime.

It is a super-3 number, since 3×141132140000433³ (a number of 43 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 141132140000397 and 141132140000406.

It is not an unprimeable number, because it can be changed into a prime (141132140000633) by changing a digit.

It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 7840674444460 + ... + 7840674444477.

Almost surely, 2^{141132140000433} is an apocalyptic number.

It is an amenable number.

141132140000433 is a deficient number, since it is larger than the sum of its proper divisors (62725395555761).

141132140000433 is a wasteful number, since it uses less digits than its factorization.

141132140000433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 15681348888943 (or 15681348888940 counting only the distinct ones).

The product of its (nonzero) digits is 3456, while the sum is 27.

Adding to 141132140000433 its reverse (334000041231141), we get a palindrome (475132181231574).

The spelling of 141132140000433 in words is "one hundred forty-one trillion, one hundred thirty-two billion, one hundred forty million, four hundred thirty-three", and thus it is an aban number.