RE: Brainsteem Mathematics Challenges: Factorial Sum
If you write them in binary:
1!=0000000001
2!=0000000010
3!=0000000110
4!=0000011000
5!=0001111000
6!=1011010000
To be a power of 2 there has to be only one 1.
You can't use the same number 3 times since then a the first one from left on would remain at it's place and there would be more ones.
After 6! they aren't made out of consecutive ones and you won't find any 2 numbers with the same "0 gaps" since they would have to be a power of 2 minus a smaler power of 2 and this isn't posible for 6! or higher. So you're not able to clean this up.
So you can only work with 5! or less.
If we look at the numbers
1!=0000000001
1!=0000000001
2!=0000000010
1!=0000000001
1!=0000000001
3!=0000000110
2!=0000000010
3!=0000000110
4!=0000011000
2!=0000000010
3!=0000000110
5!=0001111000
are the only ones working.
Not seen this way before :-)
thanks, upvoted!
The positive integer are 2,3,5 and 7 and this has a common difference of 2 without considering the first term which the the only number given in the equation above.
Here is the solution.
Where a = 2 , b = 3 and c = 5
a! + b! + c! = (21)+(321)+(5432*1)
a! + b! + c! = (2) + (6) + (120)
a! + b! + c! = 128
From equation 2^n = 128 where 128 can be converted to power of 2 which is 2^7
Therefore n = 7, a = 2, b = 3 and c = 5
Thanks for this brainstorming.