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

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

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