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

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

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30 Apr 2006, 21:56
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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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30 Apr 2006, 22:06
Ans 12:

P1-P5 are 3, 4, 7, 11, 9

Whats the source of this Question?
VP
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Re: PS - Sequence Number [#permalink]

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30 Apr 2006, 22:07
gmat_crack wrote:
In sequence P, P4 and P5 are 11 and 9 espectively. 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

A. 40

p5 = 9, p4 = 11, p3 = 7, p2 = 4, p1 = 10
so p1p2 = 40

or

p5 = 9, p4 = 11, p3 = 7, p2 = 4, p1 = 3
so p1p2 = 12
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01 May 2006, 03:19
I vote for 'A' (40) as the question is:
What is the largest possible product of P1 and P2?

Solving the question you receive P1=3 or 10, and P2 = 4.
So 12 or 40, obviously it's 40.
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01 May 2006, 09:48
Thanks guys I got it.
OA is A
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01 May 2006, 11:46
can anyone show how you solve the question?
VP
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Re: PS - Sequence Number [#permalink]

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01 May 2006, 20:03
gmat_crack wrote:
Professor wrote:
gmat_crack wrote:
In sequence P, P4 and P5 are 11 and 9 espectively. 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

A. 40
p5 = 9, p4 = 11, p3 = 7, p2 = 4, p1 = 10
so p1p2 = 40
or
p5 = 9, p4 = 11, p3 = 7, p2 = 4, p1 = 3
so p1p2 = 12

can u explain how you got these numbers?

while answering this question, i didn't think that it was this much lengthy and cumbersome. actually the following is the detail values for p1, p2 and p3 using simulation technique:

if p4+p3 is odd, then p5 = p4+p3. so p3 = 9-11 = -2
if p4+p3 is even, then p5 = (p4+p3)/2. so p3 = 18-11 = 7

if p2+p3 is odd, then p4 (when p3 = -2) = (p2+p3). So p2 = 11 +2 = 13
if p2+p3 is even, then p4 (when p3 = -2) = (p2+p3)/2. so p2 = 22+2 = 24
if p2+p3 is odd, then p4 (when p3 = 7) = (p2+p3). So p2 = 11 -7 = 4
if p2+p3 is even, then p4 (when p3 = 7) = (p2+p3)/2. so p2 = 22-7 = 15

if p2+p1 is odd, then p3 (when p3 = -2 and p2 = 13) = (p2+p1). So p1 = -2-13=-15
if p2+p1 is even, then p3 (when p3 = -2 and p2 = 13) = (p2+p1)/2. so p1 = -4-13 = -17
if p2+p1 is odd, then p3 (when p3 = -2 and p2 = 24) = (p2+p1). Not applicable (NA)
if p2+p1 is even, then p3 (when p3 = -2 and p2 = 24) = (p2+p1)/2. so p1 = -4-24 = -28
if p2+p1 is odd, then p3 (when p3 = -2 and p2 = 4) = (p2+p1). NA.
if p2+p1 is even, then p3 (when p3 = -2 and p2 = 4) = (p2+p1)/2. so p1 = -4-4 = -8
if p2+p1 is odd, then p3 (when p3 = -2 and p2 = 15) = (p2+p1). So p1 = -2-15=-17
if p2+p1 is even, then p3 (when p3 = -2 and p2 = 15) = (p2+p1)/2. so p1 = -4-15 = -19

if p2+p1 is odd, then p3 (when p3 = 7 and p2 = 13) = (p2+p1). So p1 = -6
if p2+p1 is even, then p3 (when p3 = 7 and p2 = 13) = (p2+p1)/2. so p1 = 1
if p2+p1 is odd, then p3 (when p3 = 7 and p2 = 24) = (p2+p1). So p1 = -17
if p2+p1 is even, then p3 (when p3 = 7 and p2 = 24) = (p2+p1)/2. so p1 = 14-24 = -10
if p2+p1 is odd, then p3 (when p3 = 7 and p2 = 4) = (p2+p1). So p1 = -3
if p2+p1 is even, then p3 (when p3 = 7 and p2 = 4) = (p2+p1)/2. so p1 = 10
if p2+p1 is odd, then p3 (when p3 = 7 and p2 = 15) = (p2+p1). So p1 = -8
if p2+p1 is even, then p3 (when p3 = 7 and p2 = 15) = (p2+p1)/2. so p1 = -1

there are 16 values for p1, 4 values for p2 and 2 values for p3. the highest value is still 40.
will correct, if any.
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01 May 2006, 21:07
Yep I got 40...

thanks prof you saved me lot of typing time!
01 May 2006, 21:07
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# In sequence P, P4 and P5 are 11 and 9 respectively. Each

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