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13001421412022 = 26500710706011
BaseRepresentation
bin1011110100110010000100…
…0110101101011010110110
31201000220221102010202000012
42331030201012231122312
53201003402340141042
643352434234220222
72511215225403641
oct275144106553266
951026842122005
1013001421412022
11416296549367a
12155b919884672
1373404b2c2c14
1432d3b4b0ca58
151782e435ae82
hexbd3211ad6b6

13001421412022 has 4 divisors (see below), whose sum is σ = 19502132118036. Its totient is φ = 6500710706010.

The previous prime is 13001421412019. The next prime is 13001421412061. The reversal of 13001421412022 is 22021412410031.

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

It is a super-2 number, since 2×130014214120222 (a number of 27 digits) contains 22 as substring.

It is a congruent number.

It is an unprimeable number.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3250355353004 + ... + 3250355353007.

It is an arithmetic number, because the mean of its divisors is an integer number (4875533029509).

Almost surely, 213001421412022 is an apocalyptic number.

13001421412022 is a deficient number, since it is larger than the sum of its proper divisors (6500710706014).

13001421412022 is an equidigital number, since it uses as much as digits as its factorization.

13001421412022 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6500710706013.

The product of its (nonzero) digits is 768, while the sum is 23.

Adding to 13001421412022 its reverse (22021412410031), we get a palindrome (35022833822053).

The spelling of 13001421412022 in words is "thirteen trillion, one billion, four hundred twenty-one million, four hundred twelve thousand, twenty-two".

Divisors: 1 2 6500710706011 13001421412022