if n is a positive integer greater than 8, which of the following must

Question

if n is a positive integer greater than 8, which of the following must be divisible by 4?

(A) n(n-1)(n+2)(n+3)
(B) n(n-1)(n+3)(n+4)
(C) n(n-3)(n+2)(n+4)
(D) n(n-3)(n+2)(n+3)
(E) n(n+1)(n+3)(n+4)

sowragu · Answer

All the 5 choices must be divisible by 4,if we choose n as any number divisible by 4.

For example, if I chose n as 12 then all the choices mentioned above will be divisible by 4 since '4' is a factor of '12' and for all the choices listed above 12 is the factor.

stonecold · Answer

Pick any number say 9
all the 4 numbers must be divisible by 4 else the answer will be NO
hence D

davedekoos · Answer

To be divisible by 4, a number must have 4 as a factor. Any number composed of the multiplication of two or more even numbers will have 4 as a factor. (Every even number has 2 as a factor, therefore two even numbers multiplied together will have two 2's as factors, hence 4 as a factor).

With the given answer choices, the only way we can be sure to have two even numbers is:
- if n is even, we then need one of: (n+2), (n-2), (n+4), etc. Any (n+a), where a is an even number.
- if n is odd, we then need two of: (n+1), (n-1), (n+3), etc. Any two (n+b), where b is an odd number.

In fact the only answer choice that DOESN'T meet both conditions above is answer choice C.

This can be verified by picking some numbers:

Try n=9
(A) n(n-1)(n+2)(n+3) = 9*8*11*12 --> has 4 as a factor
(B) n(n-1)(n+3)(n+4) = 9*8*12*13 --> has 4 as a factor
(C) n(n-3)(n+2)(n+4) = 9*6*11*13 --> does  NOT  have 4 as a factor
(D) n(n-3)(n+2)(n+3) = 9*6*11*12 --> has 4 as a factor
(E) n(n+1)(n+3)(n+4) = 9*10*12*13 --> has 4 as a factor

Try n=10
(A) n(n-1)(n+2)(n+3) = 10*9*12*13 --> has 4 as a factor
(B) n(n-1)(n+3)(n+4) = 10*9*13*14 --> has 4 as a factor
(C) n(n-3)(n+2)(n+4) = 10*7*12*14 --> has 4 as a factor
(D) n(n-3)(n+2)(n+3) = 10*7*12*13 --> has 4 as a factor
(E) n(n+1)(n+3)(n+4) = 10*11*13*14 --> has 4 as a factor

Try n=11
(A) n(n-1)(n+2)(n+3) = 11*10*13*14 --> has 4 as a factor
(B) n(n-1)(n+3)(n+4) = 11*10*14*15 --> has 4 as a factor
(C) n(n-3)(n+2)(n+4) = 11*8*13*15 --> has 4 as a factor
(D) n(n-3)(n+2)(n+3) = 11*8*13*14 --> has 4 as a factor
(E) n(n+1)(n+3)(n+4) = 11*12*14*15 --> has 4 as a factor

If n=12, then all options will obviously have 4 as a factor.

So the only case where 4 is not a factor is for answer C, and only when n=4x+1, where x is an integer (i.e. one more than a multiple of 4).

Either the question is written incorrectly, or it is fundamentally flawed... If the question was written as "Which of the following might not be divisible by 4?", then we would have answer C.

stonecold · Answer

We need to pick 4 numbers that are consecutive numbers example check 9,10,11,12 in all four cases and we can easily say that D is sufficient

Vinayprajapati · Answer

Agree with you. The Question seems fundamentally flawed...

unverifiedvoracity · Answer

D.

For any integer n, D is the only answer choice which will always be divisible by 4.

A product of 4 consecutive integers will always be divisible by 4. Since, none of the answer choices are a product of 4 consecutive integers, we have to ensure at least one of the integers is always a multiple of 4. The best way to do that is to ensure the answer choice is  n  *  OR   (n+3) ] *  OR   (n+2) ] *  (n-3)   OR  (n+1)].

Answer choice D is highlighted in red in the expression above.

KarishmaB · Answer

The question is flawed. Option (D) "MUST BE DIVISIBLE" by 4 for all values of n but the same is true for options (A), (B) and (E) as well. Note that you do not need only one of the numbers to be a multiple of 4. If any two numbers are even, the product will automatically become divisible by 4. In options (A), (B), (D) and (E), you have two pairs of consecutive integers each. Option (A): (n-1) and n are consecutive. So one of them will be even. (n+2) and (n+3) are consecutive. So one of them will be even too. So two of the 4 numbers will be even. Hence the product will be divisible by 4 under all circumstances. The same is true for the other specified options too. If we were talking about divisibility by 8, then we would need a multiple of 4. You use number properties discussed here in that case: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... c-or-math/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... h-part-ii/ Any 4 consecutive integers will have two even numbers one of which will be a multiple of 4. So n(n+1)(n+2)(n+3) will be divisible by 8 for all integer values of n. But such an option is not available. So we need to think some more. Considering division by 4, status of n will be equivalent to status of (n+4), (n - 4), (n + 8), (n - 8) etc. This means that if n is divisible by 4, so are all these other factors. If n is not, neither are these other factors. Similarly, (n+1) is equivalent to (n+5), (n + 9), (n - 3), (n - 7) etc. Note option (D): n(n-3)(n+2)(n+3) Here, instead of (n+1), we have (n-3) which are anyway equivalent considering division by 4. So effectively, we have 4 consecutive integers and hence option (D) will always be divisible by 8.

if n is a positive integer greater than 8, which of the following must

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if n is a positive integer greater than 8, which of the following must

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