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

(1) n^2 - 14n + 49 = 0
(2) n + 2 is the first of 3 consecutive integers whose product is 990
[Reveal] Spoiler: OA

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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, 03:14
(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

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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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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