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1320202013113 = 5922376305307
BaseRepresentation
bin10011001101100010001…
…011000000100110111001
311200012200011000001101001
4103031202023000212321
5133112233203404423
62450254210525001
7164244556645654
oct23154213004671
94605604001331
101320202013113
114699926896a5
12193a44b96761
13976575b2759
1447c8061889b
152451c796cad
hex133622c09b9

1320202013113 has 4 divisors (see below), whose sum is σ = 1342578318480. Its totient is φ = 1297825707748.

The previous prime is 1320202013111. The next prime is 1320202013149. The reversal of 1320202013113 is 3113102020231.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 1320202013113 - 21 = 1320202013111 is a prime.

It is not an unprimeable number, because it can be changed into a prime (1320202013111) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 11188152595 + ... + 11188152712.

It is an arithmetic number, because the mean of its divisors is an integer number (335644579620).

Almost surely, 21320202013113 is an apocalyptic number.

It is an amenable number.

1320202013113 is a deficient number, since it is larger than the sum of its proper divisors (22376305367).

1320202013113 is an equidigital number, since it uses as much as digits as its factorization.

1320202013113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 22376305366.

The product of its (nonzero) digits is 216, while the sum is 19.

Adding to 1320202013113 its reverse (3113102020231), we get a palindrome (4433304033344).

The spelling of 1320202013113 in words is "one trillion, three hundred twenty billion, two hundred two million, thirteen thousand, one hundred thirteen".

Divisors: 1 59 22376305307 1320202013113