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Manager  G
Joined: 15 Dec 2015
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GMAT 1: 680 Q49 V34 GPA: 4
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If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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5 00:00

Difficulty:   95% (hard)

Question Stats: 52% (02:42) correct 48% (02:28) wrong based on 92 sessions

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If p, q and r are three consecutive integers, in that order and p > 1, is their product divisible by 4?

(1) The average of p, q and r is a multiple of 2
(2) qr/p is an integer .

Originally posted by DHAR on 04 Feb 2018, 12:56.
Last edited by gmatbusters on 10 Oct 2018, 06:12, edited 1 time in total.
Edited OA
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Joined: 07 Dec 2017
Posts: 1155
Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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1
1
DHAR wrote:
If p, q and r are three consecutive integers, in that order and p > 1, is their product divisible by 4?

(1) The average of p, q and r is a multiple of 2
(2) qr/p is an integer .

As this question deals with integer properties, we'll go for a Logical approach.

Three consecutive integers can be even, odd, even in which case their product is divisible by 4 or
odd, even, odd in which case it is divisible by 4 only if the middle number (q) is divisible by 4.
We'll look for a statement which gives us this information.

(1) p + q + r = p + (p + 1) + (p + 2) = 3p +3 so the average is p + 1. Then p + 1 = q is even. But is q divisible by 4?
Insufficient!

(2) if qr = (p + 1)(p + 2) = p^2 + 3p + 2 is divisible by p then 2 must be divisible by p. Since p > 1, then p = 2.
Sufficient!

EDIT: I'm surprised that the OA is (E). Don't think I missed anything...
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Originally posted by DavidTutorexamPAL on 04 Feb 2018, 14:50.
Last edited by DavidTutorexamPAL on 05 Feb 2018, 00:15, edited 1 time in total.
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GMAT 1: 660 Q47 V34 Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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How is it Option E?

Could anyone explain?
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Joined: 22 Aug 2013
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Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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DavidTutorexamPAL wrote:
DHAR wrote:
If p, q and r are three consecutive integers, in that order and p > 1, is their product divisible by 4?

(1) The average of p, q and r is a multiple of 2
(2) qr/p is an integer .

As this question deals with integer properties, we'll go for a Logical approach.

Three consecutive integers can be even, odd, even in which case their product is divisible by 4 or odd, even, odd in which case it is not.
We'll look for a statement which gives us this information.

(1) p + q + r = p + (p + 1) + (p + 2) = 3p +3 so the average is p + 1. If this is even, then p is odd.
Sufficient!

(2) if qr = (p + 1)(p + 2) = p^2 + 3p + 2 is divisible by p then 2 must be divisible by p. Since p > 1, then p = 2.
Sufficient!

DHAR : I thought that, if sufficient, (1) and (2) were supposed to give identical answers to the question? In this case (1) gives No and (2) gives Yes... otherwise, nice question!

EDIT: I'm surprised that the OA is (E). Don't think I missed anything...

Hi

Even if the order is odd, even, odd - still the product can be divisible by 4 if the middle number is a mutliple of 4. Eg., 3, 4, 5 or 7, 8, 9.
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Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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DHAR wrote:
If p, q and r are three consecutive integers, in that order and p > 1, is their product divisible by 4?

(1) The average of p, q and r is a multiple of 2
(2) qr/p is an integer .

Let the numbers be p, p+1, p+2 respectively. Their average will be the middle number only.
Average = (p + p+1 + p+2)/3 = p+1.
We have to determined whether p*(p+1)*(p+2) is divisible by 4.

(1) Average = p+1 is even. So p is odd.
Even if p is odd, still the product p*(p+1)*(p+2) might or might not be divisible by 4.

eg, 3*4*5 is divisible by 4 but 5*6*7 is NOT divisible by 4.
So this statement is Not Sufficient.

(2) qr/p is an integer, so the product of q*r is divisible by p.
Product of (p+1)*(p+2) = (p^2 + 3p + 2) is given to be divisible by p.
Now p^2 and 3p are both multiples of p, so these two terms will anyway be divisible by p. But 2 is also divisible by p, This can only happen if p=2 and nothing else (because p>1 given).

So from this condition we can determine that the only case possible here is p, q, r as 2, 3, 4 respectively. And their product, as we can see, is divisible by 4. So this statement is Sufficient.

(OA is E, either I am missing something or request to please check the OA)
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Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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amanvermagmat wrote:
Hi

Even if the order is odd, even, odd - still the product can be divisible by 4 if the middle number is a mutliple of 4. Eg., 3, 4, 5 or 7, 8, 9.

You're right! Slipped up there... fixed my reply, thanks.
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Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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DHAR wrote:
If p, q and r are three consecutive integers, in that order and p > 1, is their product divisible by 4?

(1) The average of p, q and r is a multiple of 2
(2) qr/p is an integer .

The question is till flawed. From (1) p = odd and from (2) p = even. On the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other or the stem.
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Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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1
If p,q and r are three consecutive integers, in that order and p>1, is their product divisible by 4?

1. The average of p,q and r is a multiple of 2.

2. $$qr/p$$ is an integer
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Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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IMO C

1. (p+q+r)/3 = 2*int or p+q+r = 6*int. Consider 5,6,7 and 3,4,5 Insufficient. Another way q is the the average as p,q,r are consecutive integers. Hence q = 2 int

2. rq/p = int rq = p*int therefore pqr = p^2*int. When int is 2 answer is Yes when its 3 Answer is No.

Combining, we can get C
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GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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2
a70 wrote:
If p,q and r are three consecutive integers, in that order and p>1, is their product divisible by 4?

1. The average of p,q and r is a multiple of 2.

2. $$qr/p$$ is an integer

We know that the three integers are consecutive, and that p>1. It's a divisibility problem, so we want to start thinking about what we can deduce regarding divisibility.

When you have three consecutive integers, there are only two possibilities regarding even/odd: the integers either go EVEN-ODD-EVEN, or ODD-EVEN-ODD. Either one of those could be true in this case, given the limited information we have so far.

If the integers go EVEN-ODD-EVEN, then their product will definitely be divisible by 4. After all, p is divisible by 2, and r is divisible by 2, so pr is divisible by 4 (and therefore pqr is divisible by 4.)

If the integers go ODD-EVEN-ODD, their product MIGHT be divisible by 4. If the even integer in the middle is divisible by 4 already, then the product will also be divisible by 4. But if it isn't, the product won't be, either. For instance, 3*4*5 is divisible by 4, but 5*6*7 isn't.

Statement 1 The average of the three numbers is even. Well, they're consecutive, so their average is equal to the middle number, or q. This statement is actually just saying "q is even". If q is even, we know we're looking at the ODD-EVEN-ODD scenario - in which case we don't know whether the product is divisible by 4, as discussed above. Not sufficient.

Statement 2 qr/p is an integer. This is an interesting one. There are other ways to think about it (like case testing), but here's how I approached it.

Since the integers are consecutive, I can write everything in terms of p. The statement really says, (p+1)(p+2)/p is an integer.

Then, I simplified: (p^2 + 3p + 2)/p is an integer.

You can split this up as follows: p^2/p + 3p/p + 2/p is an integer, or in other words, p + 3 + 2/p is an integer.

I already know that p + 3 is an integer, so 2/p also has to be an integer. When is 2 divided by p an integer? Only if p is -2, -1, 1, or 2. But the first three possibilities are off the table, since the question already says that p > 1.

Therefore, this statement really tells us that p = 2. That's sufficient to answer the question. The answer is B.

---

This isn't a 'well-formed' DS question, by the way. In official DS questions, the two statements will never give contradictory information. In this one, statement 1 tells us that p is odd, but statement 2 tells us that p is equal to 2. Since those contradict, you couldn't see a DS question like this one on the test.
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Re: If p, q and r are three consecutive integers, in that order and p > 1,  [#permalink]

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DavidTutorexamPAL wrote:
DHAR wrote:
If p, q and r are three consecutive integers, in that order and p > 1, is their product divisible by 4?

(1) The average of p, q and r is a multiple of 2
(2) qr/p is an integer .

As this question deals with integer properties, we'll go for a Logical approach.

Three consecutive integers can be even, odd, even in which case their product is divisible by 4 or
odd, even, odd in which case it is divisible by 4 only if the middle number (q) is divisible by 4.
We'll look for a statement which gives us this information.

(1) p + q + r = p + (p + 1) + (p + 2) = 3p +3 so the average is p + 1. Then p + 1 = q is even. But is q divisible by 4?
Insufficient!

(2) if qr = (p + 1)(p + 2) = p^2 + 3p + 2 is divisible by p then 2 must be divisible by p. Since p > 1, then p = 2.
Sufficient!

EDIT: I'm surprised that the OA is (E). Don't think I missed anything...

--------------------------------------------------------------------------------------------------
Hi David,
Thanks for the fantastic reply. i understood the explanation but i tried it in a different way.
This is my approach
Let us take (n-1),n and (n+1) as three consecutive integers for P,Q and R.
As per second statement QR/P is an integer.
So based on the values that i have chosen for P,Q and R
i arrived at [n*(n+1)] / (n-1) = Intezer (I Assumed some value K)

Now if i go to the question stem,
whether PQR / 4 or not ??
If I go ahead and substitute the values, then i get the value of expression as [k * (n-1)^2] / 4
in this final value whether it is divisible by 4 or not entirely depends on K and n-1 value
Hence, I concluded that B is not sufficient
Please suggest me where I am going wrong in this approach Re: If p, q and r are three consecutive integers, in that order and p > 1,   [#permalink] 10 Oct 2018, 10:55
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