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20 Jan 2020, 02:58
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
59% (03:00) correct
41%
(03:00)
wrong
based on 277
sessions
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Time
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Not Attempted Yet
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
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
20 Jan 2020, 05:10
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Bunuel
Bunuel
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
General Discussion
20 Jan 2020, 05:04
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Bunuel
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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Screenshot 2020-01-20 at 5.36.05 PM.png [ 536.59 KiB | Viewed 6673 times ]
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.
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
factorize given options only option C ; 384 ; 24 * 2^4 ;
possible value
IMO C
Bunuel
