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1310433221417 is a prime number
BaseRepresentation
bin10011000100011011111…
…010000001101100101001
311122021110100022000102012
4103010123322001230221
5132432231401041132
62442001003445305
7163450520411465
oct23043372015451
94567410260365
101310433221417
1146582a432931
12191b79526835
139675b788538
14475d507d4a5
1524149d5a0b2
hex1311be81b29

1310433221417 has 2 divisors, whose sum is σ = 1310433221418. Its totient is φ = 1310433221416.

The previous prime is 1310433221401. The next prime is 1310433221429. The reversal of 1310433221417 is 7141223340131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 904028246416 + 406404975001 = 950804^2 + 637499^2 .

It is a cyclic number.

It is not a de Polignac number, because 1310433221417 - 24 = 1310433221401 is a prime.

It is a super-2 number, since 2×13104332214172 (a number of 25 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (1310433221477) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655216610708 + 655216610709.

It is an arithmetic number, because the mean of its divisors is an integer number (655216610709).

Almost surely, 21310433221417 is an apocalyptic number.

It is an amenable number.

1310433221417 is a deficient number, since it is larger than the sum of its proper divisors (1).

1310433221417 is an equidigital number, since it uses as much as digits as its factorization.

1310433221417 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 12096, while the sum is 32.

Adding to 1310433221417 its reverse (7141223340131), we get a palindrome (8451656561548).

The spelling of 1310433221417 in words is "one trillion, three hundred ten billion, four hundred thirty-three million, two hundred twenty-one thousand, four hundred seventeen".