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13200322021251 = 34400107340417
BaseRepresentation
bin1100000000010111000010…
…0000010100011110000011
31201201221100010201120212220
43000011300200110132003
53212233224414140001
644024051052422123
72531456166603021
oct300056040243603
951657303646786
1013200322021251
11422a252716262
121592389177943
13749a289157c1
14338c82bad711
1517d586019c36
hexc0170814783

13200322021251 has 4 divisors (see below), whose sum is σ = 17600429361672. Its totient is φ = 8800214680832.

The previous prime is 13200322021241. The next prime is 13200322021253. The reversal of 13200322021251 is 15212022300231.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 13200322021251 - 225 = 13200288466819 is a prime.

It is not an unprimeable number, because it can be changed into a prime (13200322021253) 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, 2200053670206 + ... + 2200053670211.

It is an arithmetic number, because the mean of its divisors is an integer number (4400107340418).

Almost surely, 213200322021251 is an apocalyptic number.

13200322021251 is a deficient number, since it is larger than the sum of its proper divisors (4400107340421).

13200322021251 is an equidigital number, since it uses as much as digits as its factorization.

13200322021251 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 4400107340420.

The product of its (nonzero) digits is 1440, while the sum is 24.

Adding to 13200322021251 its reverse (15212022300231), we get a palindrome (28412344321482).

The spelling of 13200322021251 in words is "thirteen trillion, two hundred billion, three hundred twenty-two million, twenty-one thousand, two hundred fifty-one".

Divisors: 1 3 4400107340417 13200322021251