Search a number
-
+
1388740550833 is a prime number
BaseRepresentation
bin10100001101010111011…
…000110010000010110001
311220202120201100111220001
4110031113120302002301
5140223120040111313
62541551213350001
7202222166625436
oct24152730620261
94822521314801
101388740550833
11495a639224a9
121a5192498301
13a0c5b00bc0c
144b3030bdb8d
15261ce921add
hex143576320b1

1388740550833 has 2 divisors, whose sum is σ = 1388740550834. Its totient is φ = 1388740550832.

The previous prime is 1388740550819. The next prime is 1388740550843. The reversal of 1388740550833 is 3380550478831.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1302355699264 + 86384851569 = 1141208^2 + 293913^2 .

It is an emirp because it is prime and its reverse (3380550478831) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1388740550833 - 237 = 1251301597361 is a prime.

It is not a weakly prime, because it can be changed into another prime (1388740550843) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 694370275416 + 694370275417.

It is an arithmetic number, because the mean of its divisors is an integer number (694370275417).

Almost surely, 21388740550833 is an apocalyptic number.

It is an amenable number.

1388740550833 is a deficient number, since it is larger than the sum of its proper divisors (1).

1388740550833 is an equidigital number, since it uses as much as digits as its factorization.

1388740550833 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 9676800, while the sum is 55.

The spelling of 1388740550833 in words is "one trillion, three hundred eighty-eight billion, seven hundred forty million, five hundred fifty thousand, eight hundred thirty-three".