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241440232001147 is a prime number
BaseRepresentation
bin110110111001011010101111…
…010000110100111001111011
31011122212110102220011021121022
4312321122233100310321323
5223121224043343014042
62213300040015533055
7101566324011401354
oct6671325720647173
91148773386137538
10241440232001147
116aa262556a3633
12230b489922a18b
13a4949405c2b3a
144389aaba28d2b
151cda633e7dcd2
hexdb96af434e7b

241440232001147 has 2 divisors, whose sum is σ = 241440232001148. Its totient is φ = 241440232001146.

The previous prime is 241440232001081. The next prime is 241440232001149. The reversal of 241440232001147 is 741100232044142.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-241440232001147 is a prime.

Together with 241440232001149, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (241440232001149) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (29) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 120720116000573 + 120720116000574.

It is an arithmetic number, because the mean of its divisors is an integer number (120720116000574).

Almost surely, 2241440232001147 is an apocalyptic number.

241440232001147 is a deficient number, since it is larger than the sum of its proper divisors (1).

241440232001147 is an equidigital number, since it uses as much as digits as its factorization.

241440232001147 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 43008, while the sum is 35.

Adding to 241440232001147 its reverse (741100232044142), we get a palindrome (982540464045289).

The spelling of 241440232001147 in words is "two hundred forty-one trillion, four hundred forty billion, two hundred thirty-two million, one thousand, one hundred forty-seven".