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 must = 24
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chris starr
New Member

USA
1 Posts

Posted - 09/27/2007 :  16:08:52  Show Profile  Reply with Quote
I am helping my child work on coming up with the number 24 using the follwing 4 numbers. They canbe used in any order, only once.
7, 7, -7, 3. You can use addition, subtraction, multiplication or division in the problem.
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royhaas
Moderator

USA
3059 Posts

Posted - 09/27/2007 :  17:43:37  Show Profile  Reply with Quote
7+7-(-7)+3
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tkhunny
Advanced Member

USA
1001 Posts

Posted - 09/28/2007 :  17:41:49  Show Profile  Reply with Quote
There are 51 ways to do it. They're not quite all the same. The ones on the bottom are a little trickier.

a = -7

7+(7+(3-a))
7+(3+(7-a))
3+(7+(7-a))
(7+7)+(3-a)
(7+3)+(7-a)
(3+7)+(7-a)
(7-a)+(7+3)
(7-a)+(3+7)
(3-a)+(7+7)
7+((7+3)-a)
7+((3+7)-a)
3+((7+7)-a)
7+((7-a)+3)
7+((3-a)+7)
3+((7-a)+7)
(7+(7+3))-a
(7+(3+7))-a
(3+(7+7))-a
(7+(7-a))+3
(7+(3-a))+7
(3+(7-a))+7
((7+7)+3)-a
((7+3)+7)-a
((3+7)+7)-a
((7+7)-a)+3
((7-a)+7)+3
((7+3)-a)+7
((3+7)-a)+7
((7-a)+3)+7
((3-a)+7)+7
7-(a-(7+3))
7+(7-(a-3))
7-(a-(3+7))
3-(a-(7+7))
7+(3-(a-7))
3+(7-(a-7))
(7+7)-(a-3)
(7+3)-(a-7)
(3+7)-(a-7)
(7-(a-7))+3
(7-(a-3))+7
(3-(a-7))+7
7-((a-7)-3)
7-((a-3)-7)
3-((a-7)-7)
3x(7-(7/a))
3x(7-(a/7))
3x((7/7)-a)
(7-(7/a))x3
(7-(a/7))x3
((7/7)-a)x3

Have fun.
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the_hill1962
Advanced Member

USA
1469 Posts

Posted - 11/05/2007 :  07:12:33  Show Profile  Reply with Quote
tkhunny, how did you arrive at "there are 51 ways..."? Is there a formula that is used? Did you use a computer program? Did you type all the solutions out by hand?
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tkhunny
Advanced Member

USA
1001 Posts

Posted - 11/05/2007 :  09:08:51  Show Profile  Reply with Quote
I wrote a surprisingly simple program. Here are the header notes:

Computes all possibilities of 4 numbers combined with the
4 basic arithmetic functions, with every possible association.
4 4 4 4 4 4 4 = 16384 initial construction
4 4 3 4 2 4 1 = 1536 actual possibilities
'' as used here, represents any of ('','','-','+')
1) A(B(CD))
2) (AB)(CD) - A sixth variety is identical to this one.
3) A((BC)D)
4) (A(BC))D
5) ((AB)C)D

After that, there are a couple additional simplifications:

Discarding redundant cases with only or only +
Discarding cases with division by zero
Type 5 - No zero-division possible with type 5
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