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Bunuel
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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chetan2u
Bunuel
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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Bunuel
chetan2u
Bunuel
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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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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chetan2u
Bunuel
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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chetan2u
Bunuel
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
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