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statements never contractdict each other

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raeesebrahim
 Post subject: statements never contractdict each other
 Post Posted: Mon Mar 31, 2008 12:20 am 
This question is from the book # properties, chapter 8, pg 82. the 2007 edition

If y and n are positive integers, is yn divisble by 7?

(1) n^2 -14n +49=0
(2) n +2 is the first of three consecutive integers whose product is 990

answer choices:


A
B
C
D
E

so what I did was I factored the the statement from #1 to get (n-7)(n-7)=0 so therefore n=7, and since n=7 and y is an integer, yn must be divisble by 7, so statement 1 was sufficient to answer the questions.

i took the second statement and did (n+2) (n+3) (n+4) and got 3n+9=990, and got stuck with solving statement #2, because the right answer is D.

How they got D behooves me, maybe you can help me :?
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KTsincere
 Post subject:
 Post Posted: Mon Mar 31, 2008 2:39 am 
Hey Raeesebrahim,
This is just a case of not reading the question carefully...
Statement (2) states that n+2 is the first of three consectuvie integers whose PRODUCT is 990.
You just switched PRODUCT for SUM... So your calculations would have been correct if it is said "... whose SUM is 990"
Since the product is 990 your formula should be (n+2)(n+3)(n+4)=990... Since the number is close to 1000 you should start be checking the product of consecutive integers around the number 10. Since 10*10*10=1000
Ex: 9*10*11=990
N+2=9
N=7

Hope it helps...

KT
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RonPurewal
 Post subject:
 Post Posted: Mon Mar 31, 2008 6:07 am 
Offline
ManhattanGMAT Staff


Posts: 7146
KTsincere wrote:
Hey Raeesebrahim,
This is just a case of not reading the question carefully...
Statement (2) states that n+2 is the first of three consectuvie integers whose PRODUCT is 990.
You just switched PRODUCT for SUM... So your calculations would have been correct if it is said "... whose SUM is 990"
Since the product is 990 your formula should be (n+2)(n+3)(n+4)=990... Since the number is close to 1000 you should start be checking the product of consecutive integers around the number 10. Since 10*10*10=1000
Ex: 9*10*11=990
N+2=9
N=7

Hope it helps...

KT


that's right.

if you don't think of the idea of checking consecutive integers in the neighborhood of 10*10*10 (which is also the fastest way i can think of), you can always factor 990 into primes, as you'd probably do in any other situation with products/factors/multiples:
990 = 99 * 10 = 33 * 3 * 10
= 2 * 3 * 3 * 5 * 11
once you have this factorization, it's easy to figure out the numbers, since 11 clearly must be one of them. the other prime factors - 2 * 3 * 3 * 5 - can be regrouped into 9 * 10, so there you have it: 9 * 10 * 11.
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