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1340113289197 is a prime number
BaseRepresentation
bin10011100000000100111…
…110011111111111101101
311202010001210120200021211
4103200010332133333231
5133424023000223242
62503350050354421
7165551154003364
oct23400476377755
94663053520254
101340113289197
11477380019071
1219788128ba11
13994ab794217
1448c0cc067db
1524cd5839317
hex13804f9ffed

1340113289197 has 2 divisors, whose sum is σ = 1340113289198. Its totient is φ = 1340113289196.

The previous prime is 1340113289191. The next prime is 1340113289227. The reversal of 1340113289197 is 7919823110431.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1141508475396 + 198604813801 = 1068414^2 + 445651^2 .

It is an emirp because it is prime and its reverse (7919823110431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1340113289197 - 27 = 1340113289069 is a prime.

It is a self number, because there is not a number n which added to its sum of digits gives 1340113289197.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1340113289191) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 670056644598 + 670056644599.

It is an arithmetic number, because the mean of its divisors is an integer number (670056644599).

Almost surely, 21340113289197 is an apocalyptic number.

It is an amenable number.

1340113289197 is a deficient number, since it is larger than the sum of its proper divisors (1).

1340113289197 is an equidigital number, since it uses as much as digits as its factorization.

1340113289197 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 326592, while the sum is 49.

The spelling of 1340113289197 in words is "one trillion, three hundred forty billion, one hundred thirteen million, two hundred eighty-nine thousand, one hundred ninety-seven".