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If n is an integer greater than 6, which of the following mu [#permalink]
 20 Sep 2012, 17:26
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If n is an integer greater than 6, which of the following must be divisible by 3 ? (A) n(n + 1)(n – 4) (B) n(n + 2)(n – 1) (C) n(n + 3)(n – 5) (D) n(n + 4)(n – 2) (E) n(n + 5)(n – 6)
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Last edited by Bunuel on 21 Sep 2012, 01:13, edited 1 time in total.
Edited the question.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
20 Sep 2012, 18:28
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carcass wrote: If n is an integer greater than 6, which of the following must be divisible by 3 ? (A) n(n + 1)(n – 4) (B) n(n + 2)(n – 1) (C) n(n + 3)(n – 5) (D) n(n + 4)(n – 2) (E) n(n + 5)(n – 6) I like this one.I do not find it discussed earlier on the board. OA later  Pick odd and even numbers, e.g. 7 and 8: A) (7)(8)(3) Y ; (8)(9)(5) Y B) (7)(9)(6) Y ; (8)(10)(7) N C) (7)(10)(2) N; No need to check for 8 D) (7)(11)(5) N; No need to check for 8 E) (7)(12)(1) Y; (8)(13)(2) N Only A works for both cases. A
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
21 Sep 2012, 01:13
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carcass wrote: If n is an integer greater than 6, which of the following must be divisible by 3 ?
(A) n(n + 1)(n – 4)
(B) n(n + 2)(n – 1)
(C) n(n + 3)(n – 5)
(D) n(n + 4)(n – 2)
(E) n(n + 5)(n – 6) Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3. For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers. Answer: A. Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.htmlHope it helps.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
21 Sep 2012, 06:54
I attacked the problem in this was, tell me if pron of errors n is an integers, so we can choose 7, 8, 9 and so on. Now, a number divisible by 3 the sum of number MUST be divisible by 3. 1) n (n+1)(n-4) ---> 8 * 9 * 4 ---> without perform multiplication the SUM of 8 + 9 + 4 = 21 and is divisible by 3 without reminder The rest of choices do not work if you try. thanks
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
22 Sep 2012, 11:06
Bunuel wrote: carcass wrote: If n is an integer greater than 6, which of the following must be divisible by 3 ?
Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.
For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers.
Hi Bunuel - I didn't quite follow this logic. Could you please elaborate. Thanks. Cheers
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
24 Sep 2012, 04:06
Jp27 wrote: Bunuel wrote: carcass wrote: If n is an integer greater than 6, which of the following must be divisible by 3 ?
Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.
For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers.
Hi Bunuel - I didn't quite follow this logic. Could you please elaborate. Thanks. Cheers Can you please tell me which part didn't you understand? Meanwhile check this question: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html It might help.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
24 Sep 2012, 07:56
Bunuel wrote: Can you please tell me which part didn't you understand? Meanwhile check this question: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html It might help. Thanks for your response. the link really helped. Im able to solve these problems.. thanks cheers
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
24 Sep 2012, 11:39
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Shortcut: In every set of 3 consecutive numbers, ONE of them must be divisible by 3 when we are multiplying each of the digits 1*2*3 2*3*4 3*4*5 4*5*6 (A) n(n + 1)(n – 4) The easiest is if we have something like n(n+1)(n+2) We know in this case we DEFINITELY have an expression that is divisible by 3. n=1 => 1*2*3 n=2 => 2*3*4 n=3 => 3*4*5 All are divisible by 3. Any expression must pass our 3 consecutive integer test. n(n + 1)(n – 4) n=1 => 1*2*-3 n=2 => 2*3*1 n=3 => 3*4*-1 Even if n = 4 we have: 4*5*0 = 0 which is divisible by 3. If n = 16 16*17*12 is divisible by 3. So (A) passes all the tests and one of numbers in the expression will be divisible by 3 so the whole expression when multiplied together will be divisible by 3.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
25 Sep 2012, 21:36
gmatpill wrote: Shortcut: In every set of 3 consecutive numbers, ONE of them must be divisible by 3 when we are multiplying each of the digits
1*2*3
2*3*4
3*4*5
4*5*6
(A) n(n + 1)(n – 4)
The easiest is if we have something like n(n+1)(n+2)
We know in this case we DEFINITELY have an expression that is divisible by 3.
n=1 => 1*2*3
n=2 => 2*3*4
n=3 => 3*4*5
All are divisible by 3.
Any expression must pass our 3 consecutive integer test.
n(n + 1)(n – 4)
n=1 => 1*2*-3
n=2 => 2*3*1
n=3 => 3*4*-1
Even if n = 4 we have:
4*5*0 = 0 which is divisible by 3.
If n = 16
16*17*12 is divisible by 3.
So (A) passes all the tests and one of numbers in the expression will be divisible by 3 so the whole expression when multiplied together will be divisible by 3. Thanks gmatpill - I have a basic question here -> 11 has a reminder 4 on division by 7 12 has a reminder 5 on division by 7 What is the reminder x * y divide by 7? - I can multiply reminders as long as I correct the excess amount... so 4 *5 = 20 20 - 7*2 = 6, which is the reminder when x * y / 7 - SO FAR CORRECT? Reminder when 99 / 15 give me a reminder of 9 But when i factorize the den into primes = 9 * 11 / 3 * 5 => 9/3 leaves reminder 0 and 11/5 reminder leaves 1 On multiplying the reminders, we get zero. So I should not factorize the denominator unless the denominator factor into the same primes IS THIS UNDERSTANDING CORRECT? MANY THANKS. 
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
26 Sep 2012, 00:02
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Every three consecutive positive integers will have one value divisible by 3.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
01 Oct 2012, 09:11
I tried plugging in odd and even values for n and got the answer A ... If we assume n as being either 7 , 8 or 9 we can easily see why A is the correct answer .. Because in every instance , either of the 3 numbers multiplied will be divisible by 3 and the entire product will meet the requirements.. if n =7 , then n-4 is divisible by 3 ... If n= 8 then n+1 is divisible by 3 If n= 9 , then n is divisible by 3 ...
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
30 Dec 2012, 10:14
mbaiseasy wrote: Every three consecutive positive integers will have one value divisible by 3. Beautiful solution Irene, Picturing the number line is a 15 second approach to this problem.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
08 Feb 2013, 07:14
just draw a table for each ans option plug in starting 7,8,9 etc. It should not take more than 75 s.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
08 Feb 2013, 08:56
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carcass wrote: If n is an integer greater than 6, which of the following must be divisible by 3 ?
(A) n(n + 1)(n – 4)
(B) n(n + 2)(n – 1)
(C) n(n + 3)(n – 5)
(D) n(n + 4)(n – 2)
(E) n(n + 5)(n – 6) The best way to tackle these kind of questions is by assuming values.. Since its says, n has to be greater than 6...assume the value of n to be 7....by this u will be able to eliminate options 3 and 4... Now take the value of n as 8....by this u will be able to eliminate options 2 and 5... Now we are left with only option 1...and that is the answer..
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
25 Feb 2013, 21:28
the most easy way is to pick numbers and plug in each choice. in the test room, we have no time for thinking hard way. og explantion explicitely declare this point.
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Re: If n is an integer greater than 6, which of the following mu [#permalink]
31 Mar 2013, 05:55
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Since this is a must be question it must be true for single digit nos as well and as we know any number is divisible by 3 if the sum of its digits is divisible by 3, so if we add up the digits or terms in the options we get the answer
(A) n + (n + 1) + (n – 4) = 3n - 3 ---- divisible by 3 for any n
(B) n + (n + 2) + (n – 1) = 3n + 1
(C) n + (n + 3) + (n – 5) = 3n - 2
(D) n + (n + 4) + (n – 2) = 3n + 2
(E) n + (n + 5) + (n – 6) = 3n - 1
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Re: If n is an integer greater than 6, which of the following mu
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31 Mar 2013, 05:55
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