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134432144040973 is a prime number
BaseRepresentation
bin11110100100001111101100…
…001100010010010000001101
3122121222120212020221211100211
4132210033230030102100031
5120110014012333302343
61153525135134322421
740213254564360022
oct3644175414222015
9577876766854324
10134432144040973
113991a396743a09
12130b1a25059411
135a01b87321b2c
14252a79c77b549
151081d4dddc79d
hex7a43ec31240d

134432144040973 has 2 divisors, whose sum is σ = 134432144040974. Its totient is φ = 134432144040972.

The previous prime is 134432144040949. The next prime is 134432144041043. The reversal of 134432144040973 is 379040441234431.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 133744899724969 + 687244316004 = 11564813^2 + 829002^2 .

It is an emirp because it is prime and its reverse (379040441234431) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2k-134432144040973 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (134432144041973) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 67216072020486 + 67216072020487.

It is an arithmetic number, because the mean of its divisors is an integer number (67216072020487).

Almost surely, 2134432144040973 is an apocalyptic number.

It is an amenable number.

134432144040973 is a deficient number, since it is larger than the sum of its proper divisors (1).

134432144040973 is an equidigital number, since it uses as much as digits as its factorization.

134432144040973 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 3483648, while the sum is 49.

The spelling of 134432144040973 in words is "one hundred thirty-four trillion, four hundred thirty-two billion, one hundred forty-four million, forty thousand, nine hundred seventy-three".