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Which of these must the factor of the product of four consecutive even
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05 Aug 2016, 21:02
05 Aug 2016, 21:02
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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
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
Re: Which of these must the factor of the product of four consecutive even
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05 Aug 2016, 23:20
05 Aug 2016, 23:20
Answer is D
80 is not factor.... It has to have one 5.
Sent from my Lenovo A7000-a using GMAT Club Forum mobile app
80 is not factor.... It has to have one 5.
Sent from my Lenovo A7000-a using GMAT Club Forum mobile app
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Joined: 12 Sep 2015
Posts: 6804
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Location: Canada
Re: Which of these must the factor of the product of four consecutive even
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08 Feb 2020, 08:53
08 Feb 2020, 08:53
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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
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
Answer: D
Cheers,
Brent
