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#1 MikeD

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Posted 13 June 2012 - 09:08 PM

The host posts a random three digit number and 6 other number (four random from 1 to 10 and another two from 25, 50, 75 & 100)

Then everyone else would have to use the six number to reach the three digit one using +, -, / and *. The first one to get the number wins, or the one with the closes total, and become the next host.

For example 485 with 3, 4, 6, 9, 25, 100.

4*100 = 400
3*25 = 75
400 + 75 = 475
475 + 9 = 484.

I will start:
Total to reach: 524
Number to use: 2, 3, 7, 9, 50 & 100
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#2 curr3nt

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Posted 13 June 2012 - 09:14 PM

523 = (100 * (7 - 2)) + 50 - (3 * 9)
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#3 curr3nt

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Posted 13 June 2012 - 09:24 PM

Question. Does the answer have to be an integer value?
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#4 curr3nt

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Posted 13 June 2012 - 09:26 PM

523.777.... = ((50-3)*100+(2*7))/9
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#5 MikeD

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Posted 13 June 2012 - 09:32 PM

No. Lets say rounded to the nearest 1000th.
And I forgot to put in my first post that everybody is allowed only one try per round. (So its a question of timing. Do I put my almost answer there, or do I try to get closer and hope no one in the mean time post the almost answer.)
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#6 curr3nt

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Posted 13 June 2012 - 09:51 PM

sorry...
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#7 MikeD

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Posted 13 June 2012 - 09:55 PM

my fault that I didn't say it earlier. So I will accept the 523.778 answer.
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#8 flamebirde

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Posted 14 June 2012 - 12:34 AM

50*100+3*9-7+2


Can you use the same number twice?
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#9 MikeD

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Posted 14 June 2012 - 06:15 AM

No can't use the same number twice.
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#10 TheCube

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Posted 14 June 2012 - 07:38 PM

Can you use squares and powers and such.
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And Jill just laughed at Jack as he sat there unconscious.

Jack should've known that Jill was a baddie . . .




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