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November 24th, 2014, 07:15 PM   #1
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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 TI-80 something graphics calculator.
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November 25th, 2014, 04:28 AM   #2
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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.
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November 25th, 2014, 07:51 AM   #3
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You can also use negation to increase the number of trivial solutions. 8 + 8 - (-8), for example.
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November 25th, 2014, 03:30 PM   #4
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Quote:
Originally Posted by skipjack View Post
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.
No, it isn't! Your comment is null and void. I can modify the rules. Please don't
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.
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November 25th, 2014, 03:38 PM   #5
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I have a solution with TWO 8s. But so far, I just cannot seem to come up with a non-trivial solution with THREE.

$\displaystyle (\sqrt {8+8})! $
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November 25th, 2014, 03:41 PM   #6
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Quote:
Originally Posted by Timios View Post
I have a solution with TWO 8s. But so far, I just cannot seem to come up with a non-trivial solution with THREE.

$\displaystyle (\sqrt {8+8})! $
This is a good direction. It can lead to useful ideas toward one or more
of the solutions.
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November 25th, 2014, 03:49 PM   #7
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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.
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November 25th, 2014, 03:58 PM   #8
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Quote:
Originally Posted by Timios View Post
If the rules permitted zeros, I would just add $\displaystyle 8 \times 0$ to my solution.
Hint:


$\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?
Thanks from Timios

Last edited by skipjack; November 27th, 2014 at 10:28 AM.
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November 25th, 2014, 04:17 PM   #9
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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.
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November 25th, 2014, 04:25 PM   #10
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Quote:
Originally Posted by Math Message Board tutor View Post
Your comment is null and void.
No it isn't. Just because your pedantry has backfired doesn't mean that others aren't justified in their comments.
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