54,26?

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54,26?

by sanju09 » Thu Oct 10, 2013 1:50 am
If x is a positive integer divisible by 5, then which of the following could be the value of (x - 1) (x + 2) (x - 3) (x + 4)?
A. 54,260
B. 54,261
C. 54,262
D. 54,264
E. 54,265


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by theCodeToGMAT » Thu Oct 10, 2013 2:10 am
Answer [spoiler]{D}[/spoiler]??

Steps:
X = A B C

C can be 0 or 5

Case 1: C = 0

(xx0-1)(xx0+2)(xx0-3)(xx0+4)
(xx9)(xx2)(xx7)(xx4)
(xxx8)(xx8)
(xxx4)


Case 2: C = 5

(xx5-1)(xx5+2)(xx5-3)(xx5+4)
(xx4)(xx7)(xx2)(xx9)
(xx8)(xx8)
(xxx4)
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by GMATGuruNY » Thu Oct 10, 2013 2:48 am
sanju09 wrote:If x is a positive integer divisible by 5, then which of the following could be the value of (x - 1) (x + 2) (x - 3) (x + 4)?
A. 54,260
B. 54,261
C. 54,262
D. 54,264
E. 54,265
Plug in increasing values for x and LOOK FOR A PATTERN.

If x=5, then (x - 1)(x + 2)(x - 3)(x + 4) = 4*7*2*9.
If x=10, then (x - 1)(x + 2)(x - 3)(x + 4) = 9*12*7*14.
If x=15, then (x - 1)(x + 2)(x - 3)(x + 4) = 14*17*12*19.
If x=20, then (x - 1)(x + 2)(x - 3)(x + 4) = 19*22*17*24.

In every case, the product is divisible by 12.
Thus, the correct answer choice must be a multiple of both 3 and 4.

For an integer to be a multiple of 4, its LAST TWO DIGITS must form a multiple of 4.
54,260
54,261
54,262
54,264
54,265
Eliminate every answer choice but A and D.

For an integer to be a multiple of 3, the SUM OF ITS DIGITS must be a multiple of 3.
A: 5+4+2+6+0 = 17.
Since 17 is not a multiple of 3, eliminate A.

The correct answer is D.
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by sanju09 » Thu Oct 10, 2013 3:01 am
theCodeToGMAT wrote:Answer [spoiler]{D}[/spoiler]??

Steps:
X = A B C

C can be 0 or 5

Case 1: C = 0

(xx0-1)(xx0+2)(xx0-3)(xx0+4)
(xx9)(xx2)(xx7)(xx4)
(xxx8)(xx8)
(xxx4)


Case 2: C = 5

(xx5-1)(xx5+2)(xx5-3)(xx5+4)
(xx4)(xx7)(xx2)(xx9)
(xx8)(xx8)
(xxx4)
good approach
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com