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# Which of these must the factor of the product of four consecutive even

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Joined: 12 Aug 2015
Posts: 2537
Schools: Boston U '20 (M)
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Which of these must the factor of the product of four consecutive even  [#permalink]

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05 Aug 2016, 20:02
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56% (01:59) correct 44% (02:02) wrong based on 39 sessions

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Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them

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Joined: 21 Jun 2016
Posts: 69
Location: India
Re: Which of these must the factor of the product of four consecutive even  [#permalink]

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05 Aug 2016, 22:20

80 is not factor.... It has to have one 5.

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Joined: 11 Sep 2015
Posts: 4579
GMAT 1: 770 Q49 V46
Re: Which of these must the factor of the product of four consecutive even  [#permalink]

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08 Feb 2020, 07:53
Top Contributor
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them

Let $$2k =$$ the smallest EVEN integer [all even numbers can be written as the product 2k, where K is an integer]
So, $$2k +2 =$$ the next EVEN integer
$$2k +4 =$$ the next EVEN integer
And $$2k +6 =$$ the greatest EVEN integer

So the product of these integers $$= (2k)(2k+2)(2k+4)(2k+6)$$
$$= (2k)[2(k+1)][2(k+2)][2(k+3)]$$
$$= (2)(2)(2)(2)(k)(k+1)(k+2)(k+3)$$

Important: notice that k, k+1, k+2 and k+3 are FOUR consecutive integers

Important property: There's a nice rule says: The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
Likewise, the product of any 11 consecutive integers will be divisible by 11, 10, 9, . . . 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.

Since k, k+1, k+2 and k+3 are FOUR consecutive integers, we know that one of those values is divisible by 2, and another value is divisible by 4
In other words, we can factor a 2 out of one of the integers, and we can factor of 4 out of one of the integers.
We can also factor a 3 out of one of the integers.

So our product becomes: $$(2)(2)(2)(2)(2)(4)(3)(?)(?)(?)$$

Now let's check the five answer choices...
1) 48 = (2)(2)(2)(2)(3), since there are four 2's and one 3 "hiding" in our product, we know that the product must be divisible by 48
2) 64 = (2)(2)(2)(2)(2)(2), since there are six 2's "hiding" in our product, we know that the product must be divisible by 64
3) 96 = (2)(2)(2)(2)(2)(3), since there are fix 2's and one 3 "hiding" in our product, we know that the product must be divisible by 96
4) 192 = (2)(2)(2)(2)(2)(2)(3), since there are six 2's and one 3 "hiding" in our product, we know that the product must be divisible by 192
5) 80 = (2)(2)(2)(2)(5), since we can't be certain that there's a 5 "hiding" in our product, we can't be certain that the product is divisible by 80

Cheers,
Brent
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Re: Which of these must the factor of the product of four consecutive even  [#permalink]

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08 Feb 2020, 08:09
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them

Product of four consecutive integers will be of the form = 2x*2(x+1)*2(x+2)*2(x+3) = 16*x(x+1)(x+2)(x+3); where x is an integer
x(x+1)(x+2)(x+3) will be divisible by 4*2*3 = 24
16*x(x+1)(x+2)(x+3) will be divisible by 24*16= 2^7*3

1) 48 = 2^4*3 is a factor of 2^7*3
2) 64 = 2^6 is a factor of 2^7*3
3) 96 = 2^5*3 is a factor of 2^7*3
4) 192 = 2^6*3 is a factor of 2^7*3
5) 80 = 2^5*5 is NOT a factor of 2^7*3

IMO D
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Joined: 07 Dec 2014
Posts: 1242
Re: Which of these must the factor of the product of four consecutive even  [#permalink]

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08 Feb 2020, 09:32
stonecold wrote:
Which of these must the factor of the product of four consecutive even integers:-
1) 48
2) 64
3) 96
4) 192
5) 80

A) 1,2 only
B) 2,3 only
C) 1,2,3 only
D) 1,2,3,4 only
E) All of them

product of least such sequence=2*4*6*8=384
384 divisible by 48, 64, 96, 192
D
Re: Which of these must the factor of the product of four consecutive even   [#permalink] 08 Feb 2020, 09:32
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