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

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If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 05 Dec 2017, 23:34
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A
B
C
D
E

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85% (01:42) correct 15% (01:53) wrong based on 77 sessions

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If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post Updated on: 06 Dec 2017, 04:41
(1) n^2 - 14n + 49=0
(n-7)^2=0
n=7
Therefore, product of y*n would be divisible by 7
(2) (n+2)(n+3)(n+4)=990=9*10*11
n+1=8
n=7
y*n=y*7 Hence we can with 100% confidence say that y*n is divisible by 7

Answer D

Originally posted by Alexey1989x on 06 Dec 2017, 03:14.
Last edited by Alexey1989x on 06 Dec 2017, 04:41, edited 1 time in total.
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If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 06 Dec 2017, 04:17
Bunuel wrote:
If y and n are positive integers, is yn divisible by 7?

(1) n^2 - 14n + 49
(2) n + 2 is the first of 3 consecutive integers whose product is 990



Alexey1989x

ANS would be D ..

1) \(n^2-14n+49 = 0.........(n-7)^2=0\)
So n=7
Thus yn is div by 7
Sufficient

2) n+2 is the first of 3 consecutive integers whose product is 1990
1990= 9*10*11
So n+2=9.....n=7
Thus yn is div by 7
Sufficient

D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 06 Dec 2017, 04:19
chetan2u wrote:
Bunuel wrote:
If y and n are positive integers, is yn divisible by 7?

(1) n^2 - 14n + 49
(2) n + 2 is the first of 3 consecutive integers whose product is 990



Alexey1989x

ANS would be D ..

1) \(n^2-14n+49 = (n-7)^2\)
So n=7
Thus yn is div by 7
Sufficient

2) n+2 is the first of 3 consecutive integers whose product is 1990
1990= 9*10*11
So n+2=9.....n=7
Thus yn is div by 7
Sufficient

D


The first statement reads: (1) n^2 - 14n + 49= 0. Copy/paste issued. Edited.
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If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 06 Dec 2017, 04:23
chetan2u wrote:
Bunuel wrote:
If y and n are positive integers, is yn divisible by 7?

(1) n^2 - 14n + 49
(2) n + 2 is the first of 3 consecutive integers whose product is 990



Alexey1989x

ANS would be D ..

1) \(n^2-14n+49 = 0..(n-7)^2=0\)
So n=7

Nope. (N-7)^2 = 0 implies N=7. However, the first statement i.e. n^2 - 14n + 49 is not given to be equal to 0 and hence we cannot deduce that n=7. N can still be any positive integer.

Therefore the answer to the problem is B and NOT D
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If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 06 Dec 2017, 04:44
Bunuel wrote:
chetan2u wrote:
Bunuel wrote:
If y and n are positive integers, is yn divisible by 7?

(1) n^2 - 14n + 49
(2) n + 2 is the first of 3 consecutive integers whose product is 990



Alexey1989x

ANS would be D ..

1) \(n^2-14n+49 =0.. (n-7)^2=0\)
So n=7
Thus yn is div by 7
Sufficient


2) n+2 is the first of 3 consecutive integers whose product is 1990
1990= 9*10*11
So n+2=9.....n=7
Thus yn is div by 7
Sufficient

D


The first statement reads: (1) n^2 - 14n + 49= 0. Copy/paste issued. Edited.



Thanks for clarification Bunuel, revised accordingly.
I believe that stating simple digits and signs (+/-) without mentioning the full equation can be buffling.
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Re: If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 06 Dec 2017, 06:55
Statement 1: n^2 - 14n + 49=0
(n-7)^2=0
n=7 , also y is given to be a positive integer.
Therefore, product of y*n would be divisible by 7
(2) (x)(x+1)(x+2)=990=9*10*11
x=9
n=7
y*n=y*7 Hence y*n is divisible by 7

Answer D
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Re: If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 06 Dec 2017, 15:07
chetan2u wrote:
Bunuel wrote:
If y and n are positive integers, is yn divisible by 7?

(1) n^2 - 14n + 49
(2) n + 2 is the first of 3 consecutive integers whose product is 990





ANS would be D ..

1) \(n^2-14n+49 = (n-7)^2\)
So n=7
Thus yn is div by 7
Sufficient

2) n+2 is the first of 3 consecutive integers whose product is 1990
1990= 9*10*11
So n+2=9.....n=7
Thus yn is div by 7
Sufficient

D


Hi, how did you decide which of the "n's" in statement one to make 7?
I'm confused by this step: n=7
Also, can you explain how n^2-14n+49 equals (n-7)^2?
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Re: If y and n are positive integers, is yn divisible by 7?  [#permalink]

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New post 06 Dec 2017, 17:47
BSchoolQ wrote:
chetan2u wrote:
Bunuel wrote:
If y and n are positive integers, is yn divisible by 7?

(1) n^2 - 14n + 49
(2) n + 2 is the first of 3 consecutive integers whose product is 990





ANS would be D ..

1) \(n^2-14n+49 = 0..(n-7)^2=0\)
So n=7
Thus yn is div by 7
Sufficient

2) n+2 is the first of 3 consecutive integers whose product is 1990
1990= 9*10*11
So n+2=9.....n=7
Thus yn is div by 7
Sufficient

D


Hi, how did you decide which of the "n's" in statement one to make 7?
I'm confused by this step: n=7
Also, can you explain how n^2-14n+49 equals (n-7)^2?



Hi..

1) it is given n+2 is the lowest of 3 consecutive integers.
990 can be written as only 9*10*11 if it were to be written as product of CONSECUTIVE integers and the smallest of the three is 9..
So n+2=9...n=7

2) \(n^2-14n+49=0.......n^2-2*7n+7^2=0\)..
This is in the form a^2-2ab+b^2=(a-b)^2
So (n-7)^2
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If y and n are positive integers, is yn divisible by 7? &nbs [#permalink] 06 Dec 2017, 17:47
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