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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
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Here,

If n = 14 than, 9*11*13*15*17*19 is factor of 11 and 45.
If n = 16 than, 11*13*15*17*19*21 is factor of 11, 21 and 45.
If n = 18 than, 13*15*17*19*21*23 is factor of 21 and 45.
If n = 24 than, 19*21*23*25*27*29 is factor of 21 and 45.

So, we always get 45 as factor of above given expression.

Therefore, option (C) is correct answer here.
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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
 
Bunuel wrote:
12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

If n is a positive two-digit even integer, which of the following must be a factor of \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\)?

I. 11
II. 21
III. 45

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III


 


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for the 12 Days of Christmas Competition

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­We can see that the expression (n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5) tells us that n must be a common factor of 3,5 so I think it is C.

GMATCoachBen avigutman KarishmaB Bunuel manasp35 JeffTargetTestPrep please let me know whether my method is correct :)­
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12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
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Gmatguy007 wrote:
Bunuel wrote:
12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

If n is a positive two-digit even integer, which of the following must be a factor of \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\)?

I. 11
II. 21
III. 45

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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­We can see that the expression (n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5) tells us that n must be a common factor of 3,5 so I think it is C.

GMATCoachBen avigutman KarishmaB Bunuel manasp35 JeffTargetTestPrep please let me know whether my method is correct :)­

I believe you wanted to say that the product must be a multiple of 3 and 5, not n. n does not need to be any of that. Also, the product being a multiple of 3 and 5 guarantees divisibility by 15, not by 45, so it's not clear how you got C. Also, it's not clear how you eliminated 11 and 21. Please check the solution here:

https://gmatclub.com/forum/12-days-of-c ... l#p3324689

Hope this helps.­
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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
Bunuel wrote:
Gmatguy007 wrote:
Bunuel wrote:
12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

If n is a positive two-digit even integer, which of the following must be a factor of \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\)?

I. 11
II. 21
III. 45

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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­We can see that the expression (n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5) tells us that n must be a common factor of 3,5 so I think it is C.

GMATCoachBen avigutman KarishmaB Bunuel manasp35 JeffTargetTestPrep please let me know whether my method is correct :)­

I believe you wanted to say that the product must be a multiple of 3 and 5, not n. n does not need to be any of that. Also, the product being a multiple of 3 and 5 guarantees divisibility by 15, not by 45, so it's not clear how you got C. Also, it's not clear how you eliminated 11 and 21. Please check the solution here:

https://gmatclub.com/forum/12-days-of-c ... l#p3324689

Hope this helps.­

­the expression above can be written as (n^2 - 5^2)*(n^2 - 3^2)*(n^2-1) and from this we can infer (at least that's what I thought) that both 3,5 are multiples of 15 and as a result the factor of the expression in total will also be a multiple of 15. From the given choices, the only one is (C) and that's how I chose it.  
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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
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Gmatguy007 wrote:
­the expression above can be written as (n^2 - 5^2)*(n^2 - 3^2)*(n^2-1) and from this we can infer (at least that's what I thought) that both 3,5 are multiples of 15 and as a result the factor of the expression in total will also be a multiple of 15. From the given choices, the only one is (C) and that's how I chose it.  

3 and 5 are not multiples of 15, they are factors of 15.­ Sorry, but I cannot follow your logic.
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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
Bunuel wrote:
Gmatguy007 wrote:
­the expression above can be written as (n^2 - 5^2)*(n^2 - 3^2)*(n^2-1) and from this we can infer (at least that's what I thought) that both 3,5 are multiples of 15 and as a result the factor of the expression in total will also be a multiple of 15. From the given choices, the only one is (C) and that's how I chose it.  

3 and 5 are not multiples of 15, they are factors of 15.­ Sorry, but I cannot follow your logic.

­yeah, my bad I was distracted :facepalm_man:. factor I wanted to say 
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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
Expert Reply
Gmatguy007 wrote:
Bunuel wrote:
12 Days of Christmas 🎅 GMAT Competition with Lots of Questions & Fun

If n is a positive two-digit even integer, which of the following must be a factor of \((n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5)\)?

I. 11
II. 21
III. 45

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III


 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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­We can see that the expression (n - 5)(n - 3)(n - 1)(n + 1)(n + 3)(n + 5) tells us that n must be a common factor of 3,5 so I think it is C.

GMATCoachBen avigutman KarishmaB Bunuel manasp35 JeffTargetTestPrep please let me know whether my method is correct :)­

­n can take any value e.g. it could be 14. This question is based on the concept of factors of consecutive integers. 
Say if I have 2 consecutive integers, their product will certainly have 2 as a factor because one of the numbers will be even. Similarly if I have 3 consecutive integers, their product will be divisible by 2 and 3 because at least one even number and one multiple of 3 will be there. 
Check this:
https://youtu.be/DxIH8rjhpKY

If n is two digit even integer, all the given terms will be odd. If we have any 6 consecutive odd multiples such as 1, 3, 5, 7, 9, 11, two of them will be multiples of 3 and at least one of them will be a multiple of 5. It is not necessary to have a multiple of 7 or 11 in them. Hence only 45 is acceptable. 
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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
KarishmaB wrote:
­n can take any value e.g. it could be 14. This question is based on the concept of factors of consecutive integers. 
Say if I have 2 consecutive integers, their product will certainly have 2 as a factor because one of the numbers will be even. Similarly if I have 3 consecutive integers, their product will be divisible by 2 and 3 because at least one even number and one multiple of 3 will be there. 
Check this:
https://youtu.be/DxIH8rjhpKY

If n is two digit even integer, all the given terms will be odd. If we have any 6 consecutive odd multiples such as 1, 3, 5, 7, 9, 11, two of them will be multiples of 3 and at least one of them will be a multiple of 5. It is not necessary to have a multiple of 7 or 11 in them. Hence only 45 is acceptable. 

­thanks Karishma !
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Re: 12 Days of Christmas GMAT Competition - Day 10: If n is a positive two [#permalink]
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