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 not really hard at all

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hurdler
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PostSubject: not really hard at all   Mon Aug 25, 2008 4:13 pm

easiest IMO problem I have seen (although I've only seen a few):

Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number .

Easiest AIME #15 I have seen (again, I have not seen many):

In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss delivers them in the order 1, 2, 3, 4, 5, 6, 7, 8, 9.

While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based upon the above information, how many such after-lunch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)
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FantasyLover
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PostSubject: Re: not really hard at all   Mon Aug 25, 2008 6:19 pm

Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number .

meh, latex dosent work here.

so we need to show that 21n+4 and 14n+3 are relatively prime for every natural number.

21n+4-(14n+3)=7n+1 so 7n+1 should be a largest common factor of 21n+4 and 14n+3, which is oxymoron because 21n+4=(7n+1)*3+1 and (7n+1)*2+1.

therefore 21n+4 and 14n+3 are relatively prime.

Q.E.D.
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hurdler
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PostSubject: Re: not really hard at all   Fri Aug 29, 2008 4:33 pm

you have the right idea, but your solution should say "so the GCF of 7n+1 and 14n+3 should be a largest common factor of 21n+4 and 14n+3, by the Euclidean Algorithm" instead of "so 7n+1 should be a largest common factor of 21n+4 and 14n+3"
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#H34N1
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PostSubject: Re: not really hard at all   Sat Sep 06, 2008 11:20 am

For the AIME problem, I've seen easier, but a quick run-through of a potential solution:

Case 1: the letter arrives after lunch -> \displaystyle 2^7, as possible 8 letters, of which letter 8 has already been typed.

Case 2: The letter arrives before lunch -> There are 8, 7, 6, ..., 1 possible letters left. However, note that we must omit the case of 8 letters or else we are in essence duplicating Case 1. We must also account where the 9th letter will be in the position. In sum, we have \displaystyle \sum_{i=1}^7 k\dbinom{7}{k} = 448.

Adding up, we have \displaystyle 2^7+448 = \boxed{704}.
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