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110313132000000 = 2835637614073
BaseRepresentation
bin11001000101010001000111…
…000011000000001100000000
3112110120211110211122112021120
4121011101013003000030000
5103424332243243000000
61030341053033152240
732143601150560416
oct3105210703001400
9473524424575246
10110313132000000
113217059a17a456
121045751085b680
13497263993a967
141d3528116d2b6
15cb476ebe96a0
hex6454470c0300

110313132000000 has 1008 divisors, whose sum is σ = 383178984621216. Its totient is φ = 28145664000000.

The previous prime is 110313131999891. The next prime is 110313132000083. The reversal of 110313132000000 is 231313011.

It is a Harshad number since it is a multiple of its sum of digits (15).

It is an unprimeable number.

It is a polite number, since it can be written in 111 ways as a sum of consecutive naturals, for example, 27083997964 + ... + 27084002036.

Almost surely, 2110313132000000 is an apocalyptic number.

110313132000000 is a gapful number since it is divisible by the number (10) 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 110313132000000, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (191589492310608).

110313132000000 is an abundant number, since it is smaller than the sum of its proper divisors (272865852621216).

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

110313132000000 is an frugal number, since it uses more digits than its factorization.

110313132000000 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4220 (or 4181 counting only the distinct ones).

The product of its (nonzero) digits is 54, while the sum is 15.

Adding to 110313132000000 its reverse (231313011), we get a palindrome (110313363313011).

The spelling of 110313132000000 in words is "one hundred ten trillion, three hundred thirteen billion, one hundred thirty-two million", and thus it is an aban number.