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# In sequence P, P4 and P5 are 11 and 9 respectively. Each

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VP
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In sequence P, P4 and P5 are 11 and 9 respectively. Each [#permalink]

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05 Sep 2004, 23:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In sequence P, P4 and P5 are 11 and 9
respectively. Each term after the first two terms
in sequence P is either the sum of the previous
two terms if that sum is odd, or half the sum of
the previous two terms if the sum is even. What
is the largest possible product of P1 and P2?

(A) 40
(B) 14
(C) 12
(D) 10
(E) 7
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06 Sep 2004, 00:15
Greatest product I found was actually 105. I just drew a tree with possible answers.
P1 = 8
P2 = 5
P3 = 13
P4 = 9
P5 = 11
In the above, P1*P2 would be 40 and A would be the answer.
However, another possible answer would be:
P1 = 21
P2 = 5
P3 = 13
P4 = 9
P5 = 11
Here, P1*P2 would be 105 and would actually be the real answer.
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Paul

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06 Sep 2004, 00:25
I think its C. 12 is the answer.

P1=3
P2=4
P3=7
P4=11
P5=9

Paul, note that P4=11 and P5=9 (In sequence P, P4 and P5 are 11 and 9 respectively)
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06 Sep 2004, 00:28
venksune wrote:
I think its C. 12 is the answer.

P1=3
P2=4
P3=7
P4=11
P5=9

Paul, note that P4=11 and P5=9 (In sequence P, P4 and P5 are 11 and 9 respectively)

How did I invert that?
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06 Sep 2004, 00:31
I think the P4 and P5 in the question are 11 and 9, and not the other way round.

Here's the only path to 11 and 9:

P1: 10 or 3
P2: 4
P3: 7
P4: 11
P5: 9

So, the largest possible product = 40.

Also, this assumes that we're only concerned with positive integers.
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06 Sep 2004, 00:34
Agree with 40. I just re-worked it with P5=9 and P4=11 and answer should be 40 as shown by intr3pid
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06 Sep 2004, 00:35
Perfect intr3pid. Missed on using 10. The product is 40. Haste is waste - got that!
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06 Sep 2004, 00:36
P5 is odd, so P5 = P3+P4 -> since P5 is odd, so it will be the sum of the previous two terms
9 = P3+11
P3 = -2
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06 Sep 2004, 00:43
ywilfred wrote:
P5 is odd, so P5 = P3+P4 -> since P5 is odd, so it will be the sum of the previous two terms
9 = P3+11
P3 = -2

If you examine the question carefully, it says that if the sum is odd, then the term is also odd. It doesn't say that if the term is odd, it must have come from an odd sum. In other words, if the sum is even, half of this can be an odd term.
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06 Sep 2004, 01:54
any elaborate explanation to show how you decide if the sum should be odd or even ? don't quite get the working here... thanks.
Joined: 31 Dec 1969
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06 Sep 2004, 02:03
can anyone explain how they got the numbers for the terms in the sequence ? It seems there a number of possibilites, depending of if you decide the sum to be odd or even.
Joined: 31 Dec 1969
Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
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WE: Supply Chain Management (Energy and Utilities)
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06 Sep 2004, 02:05
here's what I mean.

p5 - 9
p4 - 11

okay, p3 + p4 has to be bigger or equal than p4. So p5 is obtained by half the sum of previous two terms

p3 - 7

but from here, there are some possibilites.

if you decide p3+p2 is odd, then p2 is 4.
if you decide p3+p2 is even, then p2 is 15.

So how do you decide which is which
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06 Sep 2004, 02:55
Hi guys,

OA is 40 (choice A)

I think first you can begin by
drawing the set-up, something
like this: _ _ _ 11 9.

You are told that, by the rules of the sequence, the fifth term, 9, is found either by summing 11 and the third term or by summing the two
terms and cutting them in half. Since it is impossible to add a positive integer to 11 to get 9, we must be adding an integer to 11 to get 18.

Therefore the third term must equal 7. Your scratch paper should
now look like this: _ _ 7 11 9. Now, the second term + 7 must equal either 22 or 11. If it equaled 22, the second term would have to be 15, but then there would be no legal possible value for the first term, so the two terms must sum to 11.

Therefore, the second term is 4. Your scratch paper now looks like _ 4 7 11 9. Then by the same logic used to this point, the first term equals either 10 (to sum to 14) or 3 (to sum to 7).

The largest possible product of the first two terms is therefore 40 or answer choice (A).

from now on, I will try to post 4-5 questions everyday and I'll give the answer at the end of the day...I want to make my contribution to this website....good job intr3pid
06 Sep 2004, 02:55
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