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raj
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Post subject: what is the easy way to calculate them? Posted: Thu Apr 19, 2007 7:35 pm |
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[questions deleted because the poster did not cite the actual author of the questions]
I am not sure about answers.
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raj
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Post subject: source of these questions is Posted: Fri Apr 20, 2007 3:42 pm |
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StaceyKoprince
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Post subject: Posted: Sun Apr 22, 2007 3:44 am |
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| ManhattanGMAT Staff |
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Posts: 6077 Location: San Francisco
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Hi, Raj
Did the owners of this web site write these questions? Or did a user just post these questions on that web site? If the latter, I'm going to need to check to see whether we can legally accept this as the cited source.
(FYI these both look harder than anything you should expect to see on this test - the test questions don't usually get this high level.)
_________________ Stacey Koprince Instructor Director of Online Community ManhattanGMAT
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bangu
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Post subject: Posted: Tue Mar 18, 2008 11:40 pm |
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Answer to 1:
0. As (13+14+15+16 = 58) so their exponents will raised to same level divided by 58 will leave remainder 0.
Answer to 2:
the important numbers are the times it takes the runners to circle the track. 400 meters is 0.4 km so the first runner travels 400 metres at 4 km/hr so it takes him
0.4/4 km × hr/km = 0.1 hr.
There are 60 minutes in an hour so 0.1 hr is 6 minutes. Hence the first runner is at the starting point at 6 min, 12 min, 18 min, ...
Similarly 0.4*60/6 = 4 minutes for second runner and 0.4*60/8 = 3 minutes for third runner.
Clearly 12th minutes is the point when all of them will be again meeting at starting point.
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KTsincere
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Post subject: Posted: Tue Mar 18, 2008 11:58 pm |
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Bangu, that is a very good method however I feel it is necessary to point out that this rule is only valid if the exponent is odd.
EX: 13^6+14^6+15^6+16^6
4826809+7529536+11390625+16777216 = 40524186/58 = 698692.86
Does not equal remainder 0
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