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Hollings
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#1
Report Thread starter 16 years ago
#1
I recently came across an interewsting maths problem, which challenged you to arrange three 8s to make all of the numbers from 1 to 9, inclusive. I have got all the numbers, except for arranging three 8s to get 3.

e.g. 8 + 8 - 8 = 8
sqrt(8x8)/8 = 1

Some are easy, like the above, but others require more difficult solutions. If anyone can provide me with an answer to make three i would be extremely grateful.

I hope this challenges some other people who might wish to undertake it.

Chris
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Darkness
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#2
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#2
Using which arithmetic operators? +, -, *, / and sqrt I presume?
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The Albatross
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#3
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What about ! as in 4! = 1+2+3+4
Can u use that ?
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shiny
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#4
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If cube roots are allowed then 8/8 + cuberoot(8) = 3 ?
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Ralfskini
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#5
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(Original post by Hollings)
I recently came across an interewsting maths problem, which challenged you to arrange three 8s to make all of the numbers from 1 to 9, inclusive. I have got all the numbers, except for arranging three 8s to get 3.

e.g. 8 + 8 - 8 = 8
sqrt(8x8)/8 = 1

Some are easy, like the above, but others require more difficult solutions. If anyone can provide me with an answer to make three i would be extremely grateful.

I hope this challenges some other people who might wish to undertake it.

Chris

or if you cant use cuberoots, just square root the one for 9.
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fishpaste
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#6
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(Original post by scarlet ibis)
What about ! as in 4! = 1+2+3+4
Can u use that ?
4! = 4 x 3 x 2 x 1
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Fermat
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#7
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sqrt(8/8+8)=3
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Hollings
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#8
Report Thread starter 16 years ago
#8
Sorry, i should have mentioned you can use any operator, e.g. !, sqrt, ^, etc., but not any other numbers - so you cant use cube root, square the number and so on.

I found the answer not too long after posting this thread - ([sqrt(8+8)]!)/8 - i didnt think of just square rooting the one for 9 !!

Chris
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