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1333222013113 is a prime number
BaseRepresentation
bin10011011001101010001…
…110010111100010111001
311201110021112102120200021
4103121222032113202321
5133320414313404423
62500250154410441
7165215325062434
oct23315216274271
94643245376607
101333222013113
11474464084381
12196479430a21
1398950b67873
14487578ac81b
1524a3084995d
hex1366a3978b9

1333222013113 has 2 divisors, whose sum is σ = 1333222013114. Its totient is φ = 1333222013112.

The previous prime is 1333222013111. The next prime is 1333222013161. The reversal of 1333222013113 is 3113102223331.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 1327859819584 + 5362193529 = 1152328^2 + 73227^2 .

It is a cyclic number.

It is not a de Polignac number, because 1333222013113 - 21 = 1333222013111 is a prime.

It is a super-3 number, since 3×13332220131133 (a number of 37 digits) contains 333 as substring.

Together with 1333222013111, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (1333222013111) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 666611006556 + 666611006557.

It is an arithmetic number, because the mean of its divisors is an integer number (666611006557).

Almost surely, 21333222013113 is an apocalyptic number.

It is an amenable number.

1333222013113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1333222013113 is an equidigital number, since it uses as much as digits as its factorization.

1333222013113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1944, while the sum is 25.

Adding to 1333222013113 its reverse (3113102223331), we get a palindrome (4446324236444).

The spelling of 1333222013113 in words is "one trillion, three hundred thirty-three billion, two hundred twenty-two million, thirteen thousand, one hundred thirteen".