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33214312131000 = 233537111371915383
BaseRepresentation
bin1111000110101010011110…
…10001101010100110111000
311100121020221112001022020120
413203111033101222212320
513323140402401143000
6154350235551253240
76665440421611620
oct743251721524670
9140536845038216
1033214312131000
11a646124336870
1238851a5b1a220
13156c133454790
1482b81d08b880
153c8ea91cd6a0
hex1e354f46a9b8

33214312131000 has 1024 divisors, whose sum is σ = 139340346163200. Its totient is φ = 6361502976000.

The previous prime is 33214312130981. The next prime is 33214312131017. The reversal of 33214312131000 is 13121341233.

It is a super-2 number, since 2×332143121310002 (a number of 28 digits) contains 22 as substring.

It is a Harshad number since it is a multiple of its sum of digits (24).

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 255 ways as a sum of consecutive naturals, for example, 2159149309 + ... + 2159164691.

It is an arithmetic number, because the mean of its divisors is an integer number (136074556800).

Almost surely, 233214312131000 is an apocalyptic number.

33214312131000 is a gapful number since it is divisible by the number (30) formed by its first and last digit.

It is an amenable number.

It is a practical number, because each smaller number is the sum of distinct divisors of 33214312131000, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (69670173081600).

33214312131000 is an abundant number, since it is smaller than the sum of its proper divisors (106126034032200).

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

33214312131000 is a wasteful number, since it uses less digits than its factorization.

33214312131000 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 16157 (or 16143 counting only the distinct ones).

The product of its (nonzero) digits is 1296, while the sum is 24.

Adding to 33214312131000 its reverse (13121341233), we get a palindrome (33227433472233).

The spelling of 33214312131000 in words is "thirty-three trillion, two hundred fourteen billion, three hundred twelve million, one hundred thirty-one thousand".