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Bunuel
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If y is an even integer greater than 4, and x = y^2 - 2y, then which o 20 Jan 2020, 02:58
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C
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85% (hard)

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58% (03:01) correct 42% (03:01) wrong based on 245 sessions

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If y is an even integer greater than 4, and x = y^2 - 2y, then which of the following must be a factor of (x^2 - 8x)

I. 144
II. 216
III. 384

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


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C
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If y is an even integer greater than 4, and x = y^2 - 2y, then which o 20 Jan 2020, 05:10
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Bunuel
If y is an even integer greater than 4, and x = y^2 - 2y, then which of the following must be a factor of (x^2 - 8x)

I. 144
II. 216
III. 394

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

Bunuel
If y is an even integer greater than 4, and x = y^2 - 2y, then which of the following must be a factor of (x^2 - 8x)

I. 144
II. 216
III. 394

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

Any even integer can be assumed as "\(2n\)" for all integers \(n\)
Let \(y = 2n\)
--> \(x = y^2 - 2y = 4n^2 - 4n = 4n(n - 1)\)

So, \(x^2 - 8x = 16n^2(n - 1)^2 - 32n(n - 1)\)
= \(16n(n - 1)[n^2 - n - 2]\)
= \(16(n - 2)(n - 1)(n)(n + 1)\)
= \(16\)*(Product of 4 consecutive integers)

Note: Product of "\(x\)" consecutive integers is ALWAYS divisible by "\(x!\)"
E.g: Case 1: 1*2 or 2*3 or 3*4 are always divisible by 2!
Case 2:1*2*3 or 2*3*4 or 3*4*5 are always divisible by 3!
Case 3: 1*2*3*4*5 or 4*5*6*7*8 or 10*11*12*13*14 are always divisible by 5! and so on . . . . .

So, \(16\)*(Product of 4 consecutive integers) is always divisible by 16*4! = 16*24 = 384

Option C

Bunuel Pls correct the value 394 to 384
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If y is an even integer greater than 4, and x = y^2 - 2y, then which o 20 Jan 2020, 05:04
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Bunuel
If y is an even integer greater than 4, and x = y^2 - 2y, then which of the following must be a factor of (x^2 - 8x)

I. 144
II. 216
III. 394

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


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Bunuel

None of them can be a factor of x^2-8x for y = 6 i.e. x = 24

So I believe that options are incorrect.

Please check
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If y is an even integer greater than 4, and x = y^2 - 2y, then which o Updated on: 20 Jan 2020, 06:02
Originally posted by Archit3110 on 20 Jan 2020, 05:44.
Last edited by Archit3110 on 20 Jan 2020, 06:02, edited 1 time in total.
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solve for y = 6 , we get x= 24* 16
factorize given options only option C ; 384 ; 24 * 2^4 ;
possible value
IMO C

Bunuel
If y is an even integer greater than 4, and x = y^2 - 2y, then which of the following must be a factor of (x^2 - 8x)

I. 144
II. 216
III. 384

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


Are You Up For the Challenge: 700 Level Questions
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If y is an even integer greater than 4, and x = y^2 - 2y, then which o 20 Jan 2020, 05:58
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Dillesh4096
Bunuel
If y is an even integer greater than 4, and x = y^2 - 2y, then which of the following must be a factor of (x^2 - 8x)

I. 144
II. 216
III. 394

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

Bunuel
If y is an even integer greater than 4, and x = y^2 - 2y, then which of the following must be a factor of (x^2 - 8x)

I. 144
II. 216
III. 394

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

Any even integer can be assumed as "\(2n\)" for all integers \(n\)
Let \(y = 2n\)
--> \(x = y^2 - 2y = 4n^2 - 4n = 4n(n - 1)\)

So, \(x^2 - 8x = 16n^2(n - 1)^2 - 32n(n - 1)\)
= \(16n(n - 1)[n^2 - n - 2]\)
= \(16(n - 2)(n - 1)(n)(n + 1)\)
= \(16\)*(Product of 4 consecutive integers)

Note: Product of "\(x\)" consecutive integers is ALWAYS divisible by "\(x!\)"
E.g: Case 1: 1*2 or 2*3 or 3*4 are always divisible by 2!
Case 2:1*2*3 or 2*3*4 or 3*4*5 are always divisible by 3!
Case 3: 1*2*3*4*5 or 4*5*6*7*8 or 10*11*12*13*14 are always divisible by 5! and so on . . . . .

So, \(16\)*(Product of 4 consecutive integers) is always divisible by 16*4! = 16*24 = 384

Option C

Bunuel Pls correct the value 394 to 384
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Thank you all. Edited typo in options.
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