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If x is the product of integers a, b, c, and d, is x divisible by 128?

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If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 29 Jan 2015, 08:24
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If x is the product of integers a, b, c, and d, is x divisible by 128?

(1) a = 24

(2) a, b, c, and d are consecutive even integers

Kudos for a correct solution.

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Re: If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 29 Jan 2015, 11:15
ans B...
statement 1 not sufficient
2) statement 2 tells us they are consecutive even nos so two will be multiple of atleast 2, one of atleast 4or2^2 and one will be multiple of atleast 8 or2^3.. all four combined will be multiple of 2^6*3=128*3 so sufficient
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Re: If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 01 Feb 2015, 11:20
2
Hello Bunuel, I gave it a try:

the question is actually is abcd/128 an integer?

(1) insufficient if a is 24 then 24bcd/128 should be, by simplifying we find 3bcd/16 which can give both integer and non-integer results:

ex: if b = 3 ,c= 7, d = 2 => (3x3x7x2)/16 = non integer
b = 2, c =2, d =4 => 3x2x2x4 = integer, div by 16

(2) is sufficient as the product of any 4 consecutive even integers will be divisible by 128 (2^7 = 2x2x2x2x2x2x2)

2, 4, 6, 8 (contains the factors that together are divisible by 128)
4, 6, 8, 10 (contains the factors that together are divisible by 128)

etc,

SO B CORRECT ANSWER

Bunuel wrote:
If x is the product of integers a, b, c, and d, is x divisible by 128?

(1) a = 24

(2) a, b, c, and d are consecutive even integers

Kudos for a correct solution.

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Re: If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 01 Feb 2015, 15:07
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Bunuel wrote:
If x is the product of integers a, b, c, and d, is x divisible by 128?

(1) a = 24

(2) a, b, c, and d are consecutive even integers

Kudos for a correct solution.



Statement 1: A= 2^3*8. Therefore, we need another four 2s from the other three integers. Not sufficient.
Statement 2: A, B, C, D are consecutive even integers. Lowest possible option will be 2, 4, 6, and 8 (or their negative counterparts), which has 2^7 in it, or even 0, 2, 4, 6 (0 is even) in which case X is 0 and 0/128=0, so answer to the question is yes. Sufficient.

B.
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Re: If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 02 Feb 2015, 05:20
Bunuel wrote:
If x is the product of integers a, b, c, and d, is x divisible by 128?

(1) a = 24

(2) a, b, c, and d are consecutive even integers

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

B. While 128 may seem an arbitrary number, look a bit closer and recognize that it's a power of 2. 27=128. If you didn't see this right away, there are clues in the answer choices; 24 from statement 1 is 3x8, and 8 is 23, and in statement 2 you're given several even numbers, so your mind might see those as clues that 2 is an important number here.

Statement 1 is insufficient alone, as without any knowledge of the other three numbers you can only account for the three 2s contained within 24.

Statement 2, however is sufficient, and you might find that by looking at patterns in numbers. Consider a string of consecutive even numbers:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24....

if you're thinking in terms of "2 to a big exponent" like 128 is, you should be conscious of where you get more than one factor of 2. If you list those numbers out in terms of factors of 2, you have:

2, (2 * 2), (2 * 3), (2 * 2 * 2), (2 * 5), (2 * 2 * 3), (2 * 7), (2 * 2 * 2 * 2), (2 * 9), (2 * 2 * 5), (2 * 11), (2 * 2 * 2 * 3)

And you should see that within any stretch of four of those, you get two that contain one 2, but one that contains at least two 2s (the multiples of 4) AND one that contains at least three 2s (the multiples of 8). So at a minimum, any string of four even numbers will give you at least seven factors of 2, meaning that just knowing you have four consecutive even numbers will be sufficient.

And a note - since 0 is divisible by 128 (and every other integer other than 0), even crossing 0 with your consecutive integers will still work, as if one of those evens is 0 the product will still be divisible by 128.
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If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 01 Nov 2016, 09:26
If x is the product of integers a, b, c, and d, is x divisible by 128?

(1) a = 24
(2) a, b, c, and d are consecutive even integers
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Re: If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 01 Nov 2016, 09:30
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Re: If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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New post 15 Jan 2017, 01:37
Great Question.
Here is what i did in this one =>

We are told that a,b,c,d are all integers.
We need to see if x=a*b*c*d is divisible by 128 or not.

Statement 1->
Lets use test cases.
Case 1=>
a=24
b=1
c=1
d=1
x=24 => Clearly it is not divisible by 128.
Case 2=>
a=24
b=1
c=1
d=128
x=24*128 => divisible by 128.
Hence not sufficient.

Statement 2->
Consecutive evens =>
a=2n
b=2n+2
c=2n+4
d=2n+6


Taking the Product => x=2n(2n+2)(2n+4)(2n+6)=> 8*n(n+1)(n+2)(n+3)

Property -> Product of n consecutive integers is always divisible by n!

Hence x=8*24k For some integer k=> 128k'
Hence x must be divisible by 128.

Hence sufficient.

Hence B.

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Re: If x is the product of integers a, b, c, and d, is x divisible by 128?  [#permalink]

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