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# Two questions for you

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Intern
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07 Aug 2008, 21:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Question 1
If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum equals 350, what is n?
n=40. How so?

Question 2
What is the greatest prime factor of 4^17 - 2^28?
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Re: Two questions for you [#permalink]

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07 Aug 2008, 21:40
2 questions for ME? I feel so special.

1: "39" 7s, and "1" 77 add together is 40

2: 4^17 = 2^34
2^(34) = 2^(28)*2^6
2^(28)*2^6 - 2^(28) = 2^(28) (2^6 -1) = 2^(28) (64-1) = 2^(28)(63)= 2^(28) (7*3*3)
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Re: Two questions for you [#permalink]

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07 Aug 2008, 22:17
thanks gmatnub

but if we didn't know that n was 40, how would we go about finding it? for instance n could be 30...we could have had "2" 77s and "28" 7s.

I don't remember the other answer choices...don't know if 30 was one. I guess we'd have to try the different answer choices and use POE?
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Re: Two questions for you [#permalink]

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07 Aug 2008, 22:30
thanks gmatnub

but if we didn't know that n was 40, how would we go about finding it? for instance n could be 30...we could have had "2" 77s and "28" 7s.

I don't remember the other answer choices...don't know if 30 was one. I guess we'd have to try the different answer choices and use POE?

You have to look at the answer choices (if there isn't any other information)...

350=7m+77n
350=7(m+11n)
50=m+11n

So you see, there are different possible solutions for this eqn. (m,n)={(39,1), (28,2), (17,3), (6,4)}

There can be 40 or 30 or 20 or 10 terms. You look at the stem and pick your answer.
_________________

Is this okay?

Re: Two questions for you   [#permalink] 07 Aug 2008, 22:30
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