Search a number
-
+
12013003135 = 57343228661
BaseRepresentation
bin10110011000000011…
…11110000101111111
31011000012120210022001
423030001332011333
5144100312100020
65304012203131
7603456561520
oct131401760577
934005523261
1012013003135
115105033071
1223b31694a7
131195a682a3
1481d644847
154a498d50a
hex2cc07e17f

12013003135 has 8 divisors (see below), whose sum is σ = 16474975776. Its totient is φ = 8237487840.

The previous prime is 12013003133. The next prime is 12013003139. The reversal of 12013003135 is 53130031021.

It is a sphenic number, since it is the product of 3 distinct primes.

It is not a de Polignac number, because 12013003135 - 21 = 12013003133 is a prime.

It is a super-4 number, since 4×120130031354 (a number of 41 digits) contains 4444 as substring.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (12013003133) 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 in 7 ways as a sum of consecutive naturals, for example, 171614296 + ... + 171614365.

It is an arithmetic number, because the mean of its divisors is an integer number (2059371972).

Almost surely, 212013003135 is an apocalyptic number.

12013003135 is a deficient number, since it is larger than the sum of its proper divisors (4461972641).

12013003135 is an equidigital number, since it uses as much as digits as its factorization.

12013003135 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 343228673.

The product of its (nonzero) digits is 270, while the sum is 19.

Adding to 12013003135 its reverse (53130031021), we get a palindrome (65143034156).

The spelling of 12013003135 in words is "twelve billion, thirteen million, three thousand, one hundred thirty-five".

Divisors: 1 5 7 35 343228661 1716143305 2402600627 12013003135