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12001332032147 is a prime number
BaseRepresentation
bin1010111010100100011100…
…1000011111001010010011
31120111022112110112100101102
42232221013020133022103
53033112212000012042
641305200312055015
72346032124054113
oct256510710371223
946438473470342
1012001332032147
113907811070317
121419b3199446b
1369094a6617a3
142d6c2087b043
1515c2aed3bb32
hexaea4721f293

12001332032147 has 2 divisors, whose sum is σ = 12001332032148. Its totient is φ = 12001332032146.

The previous prime is 12001332032143. The next prime is 12001332032149. The reversal of 12001332032147 is 74123023310021.

12001332032147 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 12001332032147 - 22 = 12001332032143 is a prime.

Together with 12001332032149, 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 (12001332032143) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6000666016073 + 6000666016074.

It is an arithmetic number, because the mean of its divisors is an integer number (6000666016074).

Almost surely, 212001332032147 is an apocalyptic number.

12001332032147 is a deficient number, since it is larger than the sum of its proper divisors (1).

12001332032147 is an equidigital number, since it uses as much as digits as its factorization.

12001332032147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 6048, while the sum is 29.

Adding to 12001332032147 its reverse (74123023310021), we get a palindrome (86124355342168).

The spelling of 12001332032147 in words is "twelve trillion, one billion, three hundred thirty-two million, thirty-two thousand, one hundred forty-seven".