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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.
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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.

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.

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

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.

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?

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..

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.
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"Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3"

Why is this? Why the numbers must have different reminders?

Any positive number will take one of three forms: 3m, (3m+1) or (3m+2) i.e. either it will be divisible by 3, will leave remainder 1 or will leave remainder 2 when divided by 3. If the number takes the form 3m, the number after it is of the form 3m+1 and the one after it is of the form 3m+2.

If we have 3 consecutive numbers such as a, (a+1), (a+2), we know for sure that at least one of them is divisible by 3 since one of them will be of the form 3m. We don't know which one but one of them will be divisible by 3.

So given numbers such as (n-1)*n*(n+1), we know that the product is divisible by 3. In the given options, we don't know whether n is divisible by 3 or not. We need to look for the option which has 3 consecutive numbers i.e. in which the terms leave a remainder of 0, 1 and 2 to be able to say that the product will be divisible by 3.

Note a product such as n(n+3)(n+6). When this is divided by 3, we cannot say whether it is divisible or not because all three factors will leave the same remainder, 1. Say n = 4. Product 4*7*10. All these factors are of the form 3m+1. We don't have any 3m factor here. So we need the factors to have 3 different remainders so that one of them is of the form 3m.
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This question can be solved rather easily by TESTing VALUES, although the work itself will take a bit longer than average and it would help a great deal if you could spot the subtle Number Properties involved.

From the question stem, you can see that we're dealing with division by 3 (or the 'rule of 3', if you learned the concept that way). You don't actually have to multiply out any of the answer choices though - you just need to find the one answer that will ALWAYS have a '3' in one of its 'pieces.' The subtle Number Property I referred to at the beginning is the 'spacing out' of the terms.

(1)(2)(3) is a multiple of 3, since it's 3 times some other integers.

(5)(6)(7) is also a multiple of 3, since we can find a 3 'inside' the 6, so we have 3x2 times some other integers.

Looking at the answer choices to this question, we're clearly NOT dealing with consecutive integers, but the 'cycle' of integers is something that we can still take advantage of.

For example, we know that... When n is an integer, (n+1)(n+2)(n+3) will include a multiple of 3, since it's 3 consecutive integers (one of those 3 terms MUST be a multiple of 3, even if you don't know exactly which one it is).

You can take this same concept and 'move around' any (or all) of the pieces:

(n+1)(n+2)(n+6) will also include a multiple of 3 (that third term is 3 'more' than 'n+3').

Instead of adding a multiple of 3 to a term, you could also subtract a multiple of 3 from a term.

eg. (n-2)(n+2)(n+3) will also include a multiple of 3 (that first timer is 3 'less' than 'n+1').

The correct answer to this question subtracts a multiple of 3 from one of the terms.

All things being equal, I'd still stick to TESTing VALUES (and not approaching the prompt with math theory) - the math is easy and you can put it 'on the pad' with very little effort.

I stumbled upon your explanation while working on this OG 16 question. I found it to be quite intuitive and helpful. Sorry, for asking question on something posted so long back

I have a question on this approach (relevant part highlighted). Is it necessary that if a number is divisible 3, the sum of individual digits of its factors should be divisible by 3?

Lets take the case of 30 which is divisible by 30. If I write 30 = 2*5*3 - the sum of digits = 2+5+3 = 10 is not divisible by 3. What am I missing here?

Thanking you in advance!

carcass wrote:

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

This is not true. 8*9*4 is divisible by 3 because one of the factors is 9. So of course the product has 3 as a factor.

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.

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 ...
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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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