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12043315220147 is a prime number
BaseRepresentation
bin1010111101000000110110…
…0001101000111010110011
31120122022220100022021101022
42233100031201220322303
53034304202144021042
641340342253330055
72352046402216121
oct257201541507263
946568810267338
1012043315220147
1139235a5522902
1214260a9a9692b
136948ab553073
142d8c845a8911
1515d41a9b4cd2
hexaf40d868eb3

12043315220147 has 2 divisors, whose sum is σ = 12043315220148. Its totient is φ = 12043315220146.

The previous prime is 12043315220119. The next prime is 12043315220203. The reversal of 12043315220147 is 74102251334021.

It is a happy number.

12043315220147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 12043315220147 - 232 = 12039020252851 is a prime.

It is not a weakly prime, because it can be changed into another prime (12043315220107) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6021657610073 + 6021657610074.

It is an arithmetic number, because the mean of its divisors is an integer number (6021657610074).

Almost surely, 212043315220147 is an apocalyptic number.

12043315220147 is a deficient number, since it is larger than the sum of its proper divisors (1).

12043315220147 is an equidigital number, since it uses as much as digits as its factorization.

12043315220147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 40320, while the sum is 35.

Adding to 12043315220147 its reverse (74102251334021), we get a palindrome (86145566554168).

The spelling of 12043315220147 in words is "twelve trillion, forty-three billion, three hundred fifteen million, two hundred twenty thousand, one hundred forty-seven".