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110005532481912 = 2332347107158325033
BaseRepresentation
bin11001000000110010101000…
…101011111101100101111000
3112102111102111200212222112120
4121000302220223331211320
5103404312312303410122
61025543502144041240
732112431434034454
oct3100625053754570
9472442450788476
10110005532481912
11320620a227a7a1
1210407986303220
13494c6285489b2
141d24420a11664
15cab76a29b75c
hex640ca8afd978

110005532481912 has 256 divisors, whose sum is σ = 296014448885760. Its totient is φ = 33984305101824.

The previous prime is 110005532481883. The next prime is 110005532481991. The reversal of 110005532481912 is 219184235500011.

It is a super-2 number, since 2×1100055324819122 (a number of 29 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 in 63 ways as a sum of consecutive naturals, for example, 4394408148 + ... + 4394433180.

It is an arithmetic number, because the mean of its divisors is an integer number (1156306440960).

Almost surely, 2110005532481912 is an apocalyptic number.

110005532481912 is a gapful number since it is divisible by the number (12) 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 110005532481912, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (148007224442880).

110005532481912 is an abundant number, since it is smaller than the sum of its proper divisors (186008916403848).

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

110005532481912 is a wasteful number, since it uses less digits than its factorization.

110005532481912 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 26802 (or 26798 counting only the distinct ones).

The product of its (nonzero) digits is 86400, while the sum is 42.

Adding to 110005532481912 its reverse (219184235500011), we get a palindrome (329189767981923).

The spelling of 110005532481912 in words is "one hundred ten trillion, five billion, five hundred thirty-two million, four hundred eighty-one thousand, nine hundred twelve".