Search a number
-
+
13004333340030 = 2325316945595643
BaseRepresentation
bin1011110100111100111010…
…1010110101010101111110
31201001012110100110000021200
42331033032222311111332
53201030333313340110
643354035215040330
72511350334363654
oct275171652652576
951035410400250
1013004333340030
11416411a16a24a
1215603b0b100a6
137343c367bc46
1432d5ad74add4
15178414d046c0
hexbd3ceab557e

13004333340030 has 48 divisors (see below), whose sum is σ = 33821936806320. Its totient is φ = 3466727852544.

The previous prime is 13004333340029. The next prime is 13004333340127. The reversal of 13004333340030 is 3004333340031.

13004333340030 is a `hidden beast` number, since 130 + 0 + 433 + 33 + 40 + 0 + 30 = 666.

It is a super-2 number, since 2×130043333400302 (a number of 27 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13004333339976 and 13004333340003.

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 22512612 + ... + 23083031.

It is an arithmetic number, because the mean of its divisors is an integer number (704623683465).

Almost surely, 213004333340030 is an apocalyptic number.

13004333340030 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

13004333340030 is an abundant number, since it is smaller than the sum of its proper divisors (20817603466290).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

13004333340030 is a wasteful number, since it uses less digits than its factorization.

13004333340030 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 45598825 (or 45598822 counting only the distinct ones).

The product of its (nonzero) digits is 11664, while the sum is 27.

Adding to 13004333340030 its reverse (3004333340031), we get a palindrome (16008666680061).

Subtracting from 13004333340030 its reverse (3004333340031), we obtain a palindrome (9999999999999).

The spelling of 13004333340030 in words is "thirteen trillion, four billion, three hundred thirty-three million, three hundred forty thousand, thirty".

Divisors: 1 2 3 5 6 9 10 15 18 30 45 90 3169 6338 9507 15845 19014 28521 31690 47535 57042 95070 142605 285210 45595643 91191286 136786929 227978215 273573858 410360787 455956430 683934645 820721574 1367869290 2051803935 4103607870 144492592667 288985185334 433477778001 722462963335 866955556002 1300433334003 1444925926670 2167388890005 2600866668006 4334777780010 6502166670015 13004333340030