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143201401234147 is a prime number
BaseRepresentation
bin100000100011110110101100…
…101001101110001011100011
3200210000220211212110120102201
4200203312230221232023203
5122232202432203443042
61224321452053554031
742106650134553004
oct4043665451561343
9623026755416381
10143201401234147
11416a0411820073
1214089483645917
1361b9aac950ba9
142750db04191ab
151184ee51e01b7
hex823daca6e2e3

143201401234147 has 2 divisors, whose sum is σ = 143201401234148. Its totient is φ = 143201401234146.

The previous prime is 143201401234127. The next prime is 143201401234163. The reversal of 143201401234147 is 741432104102341.

143201401234147 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 143201401234147 - 219 = 143201400709859 is a prime.

It is not a weakly prime, because it can be changed into another prime (143201401234127) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 71600700617073 + 71600700617074.

It is an arithmetic number, because the mean of its divisors is an integer number (71600700617074).

Almost surely, 2143201401234147 is an apocalyptic number.

143201401234147 is a deficient number, since it is larger than the sum of its proper divisors (1).

143201401234147 is an equidigital number, since it uses as much as digits as its factorization.

143201401234147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 64512, while the sum is 37.

Adding to 143201401234147 its reverse (741432104102341), we get a palindrome (884633505336488).

The spelling of 143201401234147 in words is "one hundred forty-three trillion, two hundred one billion, four hundred one million, two hundred thirty-four thousand, one hundred forty-seven".