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#1 |
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"Lucan"
Dec 2006
England
2·3·13·83 Posts |
100, 75, 50, 25, 11, 1
Target 278 (Use simple arithmetic on the the six numbers) |
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#2 |
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Nov 2008
2·33·43 Posts |
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#3 |
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I quite division it
"Chris"
Feb 2005
England
31×67 Posts |
(11*25) + (100/50) + 1
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#4 |
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Oct 2007
Manchester, UK
5·271 Posts |
Is it ... the final countdown?
I assume you meant another 1 there instead of 11. 100/25 = 4 50 + 1 = 51 4 * 51 = 204 75 - 1 = 74 204 + 74 = 278 Last fiddled with by lavalamp on 2009-01-22 at 21:22 |
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#5 |
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Dec 2008
you know...around...
3×13×17 Posts |
Me gots a solution:
11*25+(100+50)/75+1 Puzzling time: < 10 mins Last fiddled with by mart_r on 2009-01-22 at 21:33 |
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#6 |
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May 2008
21078 Posts |
(75/25-1)*(100+50-11)
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#7 |
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Jun 2003
49116 Posts |
I got Flatlanders solution to the problem as posed.
If the 11 is replaced by 1 then my solution is (50 / 25) * (100 + 1) + 75 + 1 |
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#8 |
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"Lucan"
Dec 2006
England
2×3×13×83 Posts |
I don't know why my genrally trustworthy source
deemed this such a good puzzle. I have received a correction: The numbers are supposed to be 25 50 75 100 10 1 Target 278 Last fiddled with by davieddy on 2009-01-23 at 10:46 |
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#9 |
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Nov 2008
2·33·43 Posts |
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#10 |
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"Lucan"
Dec 2006
England
2·3·13·83 Posts |
Assuming you are right (and I haven't checked) well done
and the author of the problem is redeemed ![]() David Last fiddled with by davieddy on 2009-01-23 at 14:35 |
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#11 |
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Dec 2008
you know...around...
3×13×17 Posts |
Also,
(75/25-(10+1)/50)*100=278 |
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