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If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin

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If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 01:44
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If \(m = n*(n + 7)*(n + 8)\), where n is an integer, which of the following must be true?

I. m is divisible 4.
II. m is divisible by 6.
III. m is divisible by 9.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III


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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 02:23
1
when atleast 2 consecitve integers are multiplied it is always divisible by 6
IMO B is correct
we can check n=1,2,5,7
only 7 gives no for option 1 and option 3 but option 2 always true


If m=n∗(n+7)∗(n+8)m=n∗(n+7)∗(n+8), where n is an integer, which of the following must be true?

I. m is divisible 4.
II. m is divisible by 6.
III. m is divisible by 9.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 02:26
1
m = n*(n + 7)*(n + 8)
If n = Odd --> m = odd*even*odd --> It will always be divisible by 2, but NOT by 4 always
--> I. m is divisible 4. --> NO

Possible value of m = {1*8*9, 2*9*10, 3*10*11, 4*11*12, . . . . }
II. m is divisible by 6 --> YES
iii. m is divisible by 9.--> NO (Eg: 3*10*11

IMO Option B
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 04:41
1
Quote:
If m=n∗(n+7)∗(n+8), where n is an integer, which of the following must be true?

I. m is divisible 4.
II. m is divisible by 6.
III. m is divisible by 9.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III


m=n∗(n+7)∗(n+8)
n = odd = 3: 3(10)(11) = divisible by 6
n = even = 10: 10(17)(18) = divisible by 6 and 4

Ans (B)
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 08:13
1
Now m=n(n+7)(n+8)....
As far as the property of multiple etc is concerned
I. n+7 and n+8 are consecutive, so surely a multiple of 2
II. n and n+3 and n+6 will have same property when divisiblity by 3 is concerned. So n can be written as n+6. Now this is nothing but product of THREE consecutive integers. So divisible by 3

Since we are looking for MUST, these 3 integers have to be multiple of 2 and 3, so m must be multiple of 2*3=6...

I. Divisible by 4..
If n+7 is a multiple of just 2....say n is 7, so 7(7+7)(7+8)=7*14*15....NO
II. Divisible by 6
YES we have already seen above
III. Divisible by 9
Same as n is 7...7*14*15...NO

II is MUST be true.

B
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 10:26
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I. m is divisible 4......for n=3....it's not divisible
II. m is divisible by 6.......for n from 1 to 10....so on all values of m are divisible by 6
III. m is divisible by 9....for few values m is not divisible by 9


So only B is correct

OA:B
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 13:05
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We are given that m=n*(n+7)*(n+8), we are to determine which of the following must be true. We are also given that n is an integer.

The best way to test is to set n=-1, n=1, n=2, n=-2, and check which of the conditions remains true.
If n=-1, m=(-1)(6)(7)
from the above, we know that m is divisible by 6 and not divisible by 4 and 9. Since statements I and III are not always true for n=-1, then I and III are not necessarily true for all possible values of m. Since the requirement for must be true that the statement must be true for all possible values, then discard I and III.

Because the answer choices do not include none, we can conclude that m is divisible by 6 for all integers. So II. must be true for all integer values of n.

The answer must, therefore, be option B.
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 14:58
The Answer is B

To be divisible by 4 we need 2 even numbers irrespective if n is odd or even. Like n and n+2 or n+1 and n+3. This case doesnt work if n is odd.

To be divisible by 6 , we need one number divisible by 3 and one by 2.
In this case if n is odd and not divisible by 3, every n+2,N+5, n+8 is divisble by 3. If n is odd and divisible by 3, every n, n+3 is divisble by 3.
Every n+1,n+3,n+5,n+7 is divisible by 2

If n is even, either n+7(as in case if 2) will be divisible by 3 or n+8(as in case of 4)

M cannot be divisible by 3 as we will never always have two odd numbers

Therefore answer is B

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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 18 Nov 2019, 15:48
1
If m=n∗(n+7)∗(n+8), where n is an integer, which of the following must be true?

I. m is divisible 4.
II. m is divisible by 6.
III. m is divisible by 9.

--> if n= (-1), then m=(-1)*(6)*(7)=-42

(-42) is not divisible by 4 and 9.
--> That's it. We're done. We don't need to check any other values of n.
In answer choices, it says that one of I,II or III must be true.

The answer is B.
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin  [#permalink]

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New post 22 Nov 2019, 12:36
Bunuel wrote:

Competition Mode Question



If \(m = n*(n + 7)*(n + 8)\), where n is an integer, which of the following must be true?

I. m is divisible 4.
II. m is divisible by 6.
III. m is divisible by 9.

(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III


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If n = 1, we have m = 1 x 8 x 9.

If n = 2, we have m = 2 x 9 x 10.

If n = 3, we have m = 3 x 10 x 11.

We see that when n = 3, m is not divisible by 4 or 9. However, in all instances, m is divisible by 6. Indeed, m is even (since one of the consecutive factors n + 7 or n + 8 must be even) and m is also divisible by 3 (because either n, n + 1 or n + 2 is divisible by 3 and we can write m = n * (n + 1 + 6) * (n + 2 + 6)). Thus, m is divisible by 6 regardless of what value we pick for n.

Answer: B
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Re: If m = n*(n + 7)*(n + 8), where n is an integer, which of the followin   [#permalink] 22 Nov 2019, 12:36
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