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

Question

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

HMC · Answer

Answer is D

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

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BrentGMATPrepNow · Answer

Let  2k =  the smallest EVEN 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)(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

Kinshook · Answer

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

gracie · Answer

product of least such sequence=2*4*6*8=384
384 divisible by 48, 64, 96, 192
D

RanonBanerjee · Answer

If we factorize the options 1) 2^4×3 2) 2^6 3) 2^5×3 4) 2^6×3 5) 2^4×5.

Now if we take any 4 consecutive even integers, say 12,14, 16, 18>> these lot don't have anything that contains a 5. But 3 is present. Similarly we can take any such 2-digit and 3-digit series, we will find that 3 is a must along with 2 but 5 is not.

Hence eliminate option 5.

Please let me know if this is approach is correct?

Posted from my mobile device

bumpbot · Answer

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

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

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