This is Just Perfect

They are the list of perfect numbers - the numbers whose factors (except itself) add to itself. The factors of 6 are 1, 2 and 3: $1 + 2 + 3 = 6$ The next perfect number is 8128.

This is mostly a pattern finding question;

During the sequence the units go up each time for example

6 ----> 28 ----> 496 ----> ????:

So:

Units ---> Tenths ----> Hundreds ------> Thousands

So we know that the next number has to be in the thousands but there are two possible answers, so which one is it?

Well if you look closely you'll see a pattern in the number itself for example it goes 6, 28 then 496 noticed that the number 6 repeated itself making it so that the ending of the next number will be an 8 if it continues to follow the sequence and 8128 would be this number that satisfies it.

$6\equiv 1\quad (mod\quad 5)\\ 28\equiv 3\quad (mod\quad 5)\\ 496\equiv 1\quad (mod\quad 5)$

We get the series $1,3,1,3,1,3,1,3...$

Thus the next number would be $\equiv 3\quad (mod\quad 5)$ and only 8128 satisfies this.

My solution is so basic (almost embarrassing)

6 -> 2^1 * 3

28 -> 2^2 * 7

496 -> 2^3* 62

there's only one number in the options multiple of 2^4

well look

3=2^2-1

7=2^3-1

31=2^5-1

the given series 6=((2^2-1) 2 ), 28=(2^3-1) (2^2),496=(2^5-1)*(2^4) we can arrange it like this

(2^2-1) (2)=(3) (2)=6

(2^3-1) (2 ^2)=(7) (4)=28

(2^4-1) (2 ^3)=(15) (8)=120

(2^5-1) (2 ^4)=(31) (16)=498

(2^6-1) (2 ^5)=(63) (32)=2016

(2^7-1) (2 ^6)=(127) (64)=8128

actually i just hit the answer and i got a point

The theorem is that all even perfect numbers can be predicted using the following: (2^p - 1)(2^(p-1)) where p is any prime number.

Substituting primes: (2^2 -1)(2^(2-1)) = 6.
(2^3 -1)(2^(3-1))= 28. (2^5 -1)(2^(5-1))= 496 , finally, (2^7 -1)(2^(7-1)) = 8128.

I forgot which mathematician did this. I think it was Euler or Fermat.

Just see at the last digit of the number ^o^ 6, ...8, ...6, then 8 again

I've seen a different solution... All terms follow this pattern :

G (n)=(2^n)*[2^(n+1)-1]

Only with the enunciated :

F (n)=G (2^(n-1))=2^[2^(n-1)]*{2^[2^(n-1)+1]-1}

F (1)=2^1 (2^2-1)=2 3=6

F (2)=2^2 (2^3-1)=4 7=28

F (3)=2^4 (2^5-1)=16 31=496

It was really easy I guess

Others: big solution with logic

Me: ends in 6, ends in 28, ends in 7, ends in 28!!

The one's place is always going in a pattern of 6,8,6 then only the number 8128 has in the one's place 8

For me it was like just reading the unit digits. First 6 then 8, then 6 so it follows the next should end with 8. and since there was just one option ending with 8 so Voila!

Honestly, I just guessed. I saw that the numbers were increasing in the number of digits (1 digit, 2 digits, 3 digits) and concluded that the next number would have four digits. This brought my answer choices down to two. Then, I saw that the one's place went from 6 to 8 and back to 6 and guessed that the next number would end in 8. This led me to choose the correct answer. Though it is really cool reading about the perfect numbers, etc.

They are all numbers which are the product of a certain n-power of 2 and another number which is 1 less than twice the first number.

In other words, $2^{n}$ ( $2 \times 2^{n}$ - 1)

6=2*3

28=4*7

496=16*31

Although I'm not sure why 8*15 is not in this sequence.

Its a type of number sequence. Its called Perfect Numbers. The question itself says "This is Just Perfect".

The first perfect number is 6, because 1, 2, and 3 are its proper positive divisors, and 1 + 2 + 3 = 6. Equivalently, the number 6 is equal to half the sum of all its positive divisors: ( 1 + 2 + 3 + 6 ) / 2 = 6.

The next perfect number is 28 = 1 + 2 + 4 + 7 + 14.

This is followed by the perfect numbers 496 and 8128.

source: wikipedia.

((2^n)-1)(2^n-1) where n is prime number n=2,3,5,7........... .

It was a guess...it needed to be even and to jump to a 6 figure number seemed excessive

I think, because the first number is 6 and the next number ending in 8, the third number ending in 6 again. so the next number ending in 8 ... heheheh: D