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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

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.

Re: If n is an integer greater than 6, which of the following mu [#permalink]

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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.

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?

Re: If n is an integer greater than 6, which of the following mu [#permalink]

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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.

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.

Re: If n is an integer greater than 6, which of the following mu [#permalink]

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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.

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

Re: If n is an integer greater than 6, which of the following mu [#permalink]

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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 ...
_________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

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..

Re: If n is an integer greater than 6, which of the following mu [#permalink]

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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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visit my facebook to help me. on facebook, my name is: thang thang thang

Re: If n is an integer greater than 6, which of the following mu [#permalink]

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31 Mar 2013, 05:55

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Since this is a must be true question it must be true for any n 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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Last edited by prasun9 on 08 Oct 2013, 05:38, edited 2 times in total.

We need to check only for 3 values of n because the answer will be the same for every 3 consecutive integers. Actually we need not consider one of those values as it is always a multiple of 3.

So we can take the values of n as 7 and 8 or 10 and 11 or 13 and 14 etc

If we plug in the values 7 and 8 we get the answer once we check A as it satisfies both the values. We need not actually check other choices as they would fail for either 7 or 8.
_________________

Which method is better in questions like these to apply on the test? to pick the numbers or to apply concept of divisibility by 3 ( Remainder 0,1,2)

In my opinion, using logic is almost always better than plugging in. Plugging in is full of possibilities of making mistakes - incorrect calculation, not considering all possibilities, getting lost in the options etc.

Logic is far cleaner.

I know that talking about positive integers, in any set of 3 consecutive positive integers, one integer will be divisible by 3 and the other 2 will not be.

So I am looking for 3 consecutive positive integers e.g. (n-1)n(n+1)

Note that (n-4) is equivalent to (n-1) since if (n-4) is divisible by 3, so is (n-1). If (n-4) is not divisible by 3, neither is (n-1) (because the difference between these two integers is 3) Hence (A) is equivalent to 3 consecutive integers. Answer (A)

On same lines, note that n is equivalent to (n-3), (n + 3), (n + 6) etc.
_________________

In my opinion, using logic is almost always better than plugging in. Plugging in is full of possibilities of making mistakes - incorrect calculation, not considering all possibilities, getting lost in the options etc.

Logic is far cleaner.

I know that talking about positive integers, in any set of 3 consecutive positive integers, one integer will be divisible by 3 and the other 2 will not be.

So I am looking for 3 consecutive positive integers e.g. (n-1)n(n+1)

Note that (n-4) is equivalent to (n-1) since if (n-4) is divisible by 3, so is (n-1). If (n-4) is not divisible by 3, neither is (n-1) (because the difference between these two integers is 3) Hence (A) is equivalent to 3 consecutive integers. Answer (A)

On same lines, note that n is equivalent to (n-3), (n + 3), (n + 6) etc.

Responding to a pm:

As said here, I prefer to use logic rather than number plugging. Number plugging is usually the last resort. Given above is the logical solution to this problem. For more on this topic, check:

I wish someone would solve this question step by step.

That's the point - there are no steps here. There is no process you have to follow. You just need to understand the logic and you will have your answer.