33550336 = 212*8191 So its factors are of the form 2m and 2n*8191 where 0 ≤ m,n ≤ 12. Their sum is (1 + 8191)*(20 + 21 + 22 + ... + 212) = 8192*(213 - 1)/(2 -1) = 67100672 This includes the factor 33550336 itself, so the sum of the proper factors is 67100672 - 33550336 = 33550336. Therefore the number is perfect.
The fifth perfect number is 33,550,336.
2,4,8 (1 and 16 are divisors too, but the number and 1 are always divisors)
Sum of divisors = 91.
12 has six divisors.
33550336 = 212*8191 So its factors are of the form 2m and 2n*8191 where 0 ≤ m,n ≤ 12. Their sum is (1 + 8191)*(20 + 21 + 22 + ... + 212) = 8192*(213 - 1)/(2 -1) = 67100672 This includes the factor 33550336 itself, so the sum of the proper factors is 67100672 - 33550336 = 33550336. Therefore the number is perfect.
A number is considered perfect if it is equal to the sum of all its positive factors/divisors, excluding itself. These are the first few perfect numbers: * 6 * 28 * 496 * 8128 * 33550336 * 8589869056A perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. An example 1+2+3=6 and 1x2x3=6
It is 33550336
Oh, dude, the factors of 33550336 are like those numbers that you can multiply together to get 33550336. So, it's 1, 2, 4, 8, 16, 32, 64, 131, 262, 524, 1048, 2096, 4192, 8384, 16768, 33536, 67072, 134144, 268288, 536576, 1073152, 2146304, 4292608, 8585216, 17170432, and 33550336. But, like, who's counting, right?
Yes, there is. It is 33550336.
yes. 33550336 is in fact a perfect number. i know this because on a math problem it asks "show that each number is perfect", so it must be perfect. :) ur welcome!
The fifth perfect number is 33,550,336.
6, 28, 496, 8128, 33550336
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16382, 32764, 65528, 131056, 262112, 524224, 1048448, 2096896, 4193792, 8387584, 16775168, 33550336
Divisors are used to divide numbers.
2,4,8 (1 and 16 are divisors too, but the number and 1 are always divisors)
Sum of divisors = 91.