In an alternative universe where the following is true:
1 # 1 = 3
2 # 2 = 7
1 # 2 = 5
What would ? be?
5 # 5 = ?
YOU DO NOT NEED THE FOLLOWING TO FINISH.
Extra equations:
3 # 3 = 11, 4 # 4 = 15
Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. It only takes a minute to sign up.
Sign up to join this communityIn an alternative universe where the following is true:
1 # 1 = 3
2 # 2 = 7
1 # 2 = 5
What would ? be?
5 # 5 = ?
YOU DO NOT NEED THE FOLLOWING TO FINISH.
Extra equations:
3 # 3 = 11, 4 # 4 = 15
By the way, as an alternative let's say your plus is $+^{\prime}$, now we can define the operator: $x+^{\prime}y = 2(x+y)-1$
I think the answer is clearly 17.
My Logic is that for each number digit you replace it with the next number in the fibinaci sequence, add as usual, then add by one. so we get
1+1+1=3 (second number in fibonacci is 1)
3+3+1=7
1+3+1=5
and thus 8+8+1=17
Or maybe the answer is 23! because the formula is $(x+y) + (x+y-1)$, but of course in this universe were uses the octal numbering system!
Or maybe they use a base 10 numbering system, but they order their numbers from 0-9 like this:
8,3,1,2,0,5,7,6,9,4
In which case the answer is 34!
All are correct given the 3 equations which are the 'only ones I need' to solve the puzzle :)
Without looking at the "extra equations", I have come up with this solution:
Evaluate the sum and look for the corresponding prime number.
The first few prime numbers are 2 3 5 7 11 13 17 19 23 29.
1 + 1 = 2nd prime number = 3
2 + 2 = 4th prime number = 7
2 + 1 = 3rd prime number = 5
Therefore 5 + 5 = 10th prime number = 29
+
to obfuscate the fact that they're asking you to find an unknown operator. $\endgroup$ – R.. GitHub STOP HELPING ICE Oct 11 '14 at 1:23