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32332002440147 is a prime number
BaseRepresentation
bin1110101100111111000011…
…00001010011001111010011
311020110220112211201001001022
413112133201201103033103
513214211420111041042
6152433041124430055
76544624106152454
oct726374141231723
9136426484631038
1032332002440147
11a335a1a536083
1237621aa24832b
131506b829b7227
147dac3d31912b
153b1069c84cd2
hex1d67e18533d3

32332002440147 has 2 divisors, whose sum is σ = 32332002440148. Its totient is φ = 32332002440146.

The previous prime is 32332002440071. The next prime is 32332002440149. The reversal of 32332002440147 is 74104420023323.

It is a strong prime.

It is an emirp because it is prime and its reverse (74104420023323) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-32332002440147 is a prime.

Together with 32332002440149, 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 (32332002440149) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 16166001220073 + 16166001220074.

It is an arithmetic number, because the mean of its divisors is an integer number (16166001220074).

Almost surely, 232332002440147 is an apocalyptic number.

32332002440147 is a deficient number, since it is larger than the sum of its proper divisors (1).

32332002440147 is an equidigital number, since it uses as much as digits as its factorization.

32332002440147 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 96768, while the sum is 35.

The spelling of 32332002440147 in words is "thirty-two trillion, three hundred thirty-two billion, two million, four hundred forty thousand, one hundred forty-seven".