November 24th, 2014, 07:15 PM  #1 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  3 8's to make 24  my challenge puzzle
In each expression: Use exactly three 8's. Use at most three square root symbols. If you can't access the square root symbol, then you can type "sqrt" as in the form "sqrtX" or "sqrt(X)," where appropriate. Here, "X" might be one or more characters in length. Use at most two factorial signs. You can use some combination of addition, subtraction/negation, multiplication, and/or division signs (+, , *, /). You can use parentheses. One pair should be sufficient. Concatenation is not allowed. No other numbers, operations, or characters are allowed. To start off with, I am giving you this solution: 8 + 8 + 8 > > > Try to come up with as many as five other solutions. One of the ways to test your expression might be to enter it into a TI80 something graphics calculator. 
November 25th, 2014, 04:28 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,965 Thanks: 2214 
24 = (8 + 8 + 8) = (8) + 8 + 8 = 8 + (8) + 8 = 8 + 8 + (8) = 8 + (8 + 8) = (8 + 8) + 8 You may have wanted more inventive answers, but it's too late to change the rules now. 
November 25th, 2014, 07:51 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
You can also use negation to increase the number of trivial solutions. 8 + 8  (8), for example.

November 25th, 2014, 03:30 PM  #4  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
make the error now or in the future that it is too late to change the rules. Modification of the rules:  The expression isn't valid if at least one operation, or set of parentheses, or at least both can still be removed and the expression still be equal to 24.  
November 25th, 2014, 03:38 PM  #5 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 392 Thanks: 71 
I have a solution with TWO 8s. But so far, I just cannot seem to come up with a nontrivial solution with THREE. $\displaystyle (\sqrt {8+8})! $ 
November 25th, 2014, 03:41 PM  #6 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  
November 25th, 2014, 03:49 PM  #7 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 392 Thanks: 71 
If the rules permitted zeros, I would just add $\displaystyle 8 \times 0$ to my solution.
Last edited by skipjack; November 27th, 2014 at 10:27 AM. 
November 25th, 2014, 03:58 PM  #8  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
$\displaystyle \sqrt {8+8} $ is equal to 4, correct? What if you were to subtract that expression from the third 8? What could you do with that? Last edited by skipjack; November 27th, 2014 at 10:28 AM.  
November 25th, 2014, 04:17 PM  #9 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 392 Thanks: 71 
Now I have a solution with FOUR 8s. First TWO and now FOUR. I just can't seem to hit that happy medium of THREE! $\displaystyle 8 \times \sqrt {8+ \frac {8}{8}}$ Last edited by skipjack; November 27th, 2014 at 11:51 AM. 
November 25th, 2014, 04:25 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra  

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