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chris starr
New Member
USA
1 Posts |
 Posted - 09/27/2007 : 16:08:52
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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
3044 Posts |
Posted - 09/27/2007 : 17:43:37
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| 7+7-(-7)+3 |
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tkhunny
Advanced Member
USA
1001 Posts |
Posted - 09/28/2007 : 17:41:49
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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
1444 Posts |
Posted - 11/05/2007 : 07:12:33
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| 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
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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×(C×D))
© 2) (A×B)×(C×D) - A sixth variety is identical to this one.
© 3) A×((B×C)×D)
© 4) (A×(B×C))×D
© 5) ((A×B)×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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Interactive Math Goodies Software
must = 24
