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If p and q are two consecutive positive integers and pq= 30x

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If p and q are two consecutive positive integers and pq= 30x  [#permalink]

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New post Updated on: 12 Oct 2012, 01:10
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If p and q are two consecutive positive integers and pq= 30x , is x an integer?

(1) p^2 is divisible by 25.
(2) 63 is a factor of q^2

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Originally posted by subirh on 11 Oct 2012, 20:26.
Last edited by Bunuel on 12 Oct 2012, 01:10, edited 1 time in total.
Renamed the topic and edited the question.
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Re: DS question  [#permalink]

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New post 12 Oct 2012, 00:00
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subirh wrote:
If p and q are two consecutive positive integers and pq= 30x , is x an integer?
1)p^2 is divisible by 25.
2)63 is a factor of q^2


\(q = p + 1.\)
The question can be reworded as "Is \(p(p+1)\) divisible by 30?"
The product of two consecutive integers is always divisible by 2. In order to be divisible by 30, the number should be also divisible by 3 and 5.

(1) \(p^2\) divisible by \(5^2\) implies that \(p\) divisible by 5. But we don't know anything regarding the divisibility by 3.
Not sufficient.

(2) \(63=7\cdot{9}\) is a factor of \((p+1)^2,\) so 9 is a factor of \((p+1)^2,\) therefore 3 is a factor of \(p+1\).
Again, not sufficient, we don't know whether \(p\) or \(p+1\) is divisible by 5.

(1) and (2) together: we are ensured that \(p\) is divisible by 3 and \(p+1\) is divisible by 5.
Sufficient.

Answer C
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Re: DS question  [#permalink]

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New post 11 Oct 2012, 20:52
1 --> p^2 is divisible by 25 which gives p is a divisible by 5 so p can be 5,10,15 and so on
We have no info abt q

if p=5;q=6 ==> then x=1 integer but if p=10;q=11 ==> then x is not an integer.. Not sufficient

2)63 is a factor of q^2

then 3 is a factor of q but we have no info abt p

Taken together...
5 is a factor of p and 3 is a factor of q.. so possible values are 5*6.. 20*21 and in all these cases x would be a inetger

so ans is C
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Re: If p and q are two consecutive positive integers and pq= 30x  [#permalink]

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New post 06 Sep 2016, 05:19
subirh wrote:
If p and q are two consecutive positive integers and pq= 30x , is x an integer?

(1) p^2 is divisible by 25.
(2) 63 is a factor of q^2

This question is marked as 95% Hard. So, needless to say that this is a Level 750+ question and hence there is no fastest way to solve it. Anyways, following solution is, in one way or other, what everyone above said but I am going to explain it in best way possible.

Universally Available Statements:

i) p and q are two consecutive positive integers
i. a) p > = 1; q > = 1
i. b) p - q = 1 (We do not know which one is bigger and it does not matter)

ii) p*q = 30*x = 2 * 3 * 5 * x.

Question: Is x an Integer?

How to Go about this question: If, Integer x Integer = Integer

How can x be a Non-Integer?
Possible values of x = 1, 2, 3, \(\frac{1}{2}, \frac{2}{3},\frac{3}{5}, \frac{3}{7}, \frac{7}{30}\) and so on such that given conditions are satisfied.

Solution: If 2, 3 & 5 are factors of p, q, or both, then x would always be an Integer. Hence, the question boils down to determining whether p or q are multiples of 2, 3, and 5.

=> p & q are Consecutive Integers. Therefore, their multiple MUST BE EVEN. Hence, either of p or q is a Multiple of 2.

Now we need to determine whether the given statements are sufficient to determine whether p or q are multiples of 3 & 5.

Statement 1: \(p^2\)is divisible by 25

=> p is a Multiple of 5 - Insufficient because we don't know whether 3 is a factor of p or q.

Statement 2: 63 is a factor of q^2

=> q = 3*\(\sqrt{7}\)*z (z is some variable which is a factor of x)

Since q is an Integer,

q = 21*z - Insufficient because we don't know whether 5 is a factor of p or q.
=> q is a Multiple of 3 & 7 - Insufficient because we don't know whether 5 is a factor of p or q.

Statement 1& 2 together:

p is a Multiple of 5 And q is a Multiple of 3 & 7.

=> p*q = Multiple of 2, 3, 5 and 7.
Hence, x will always be an Integer.

Option C - Sufficient.
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Re: If p and q are two consecutive positive integers and pq= 30x  [#permalink]

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New post 13 Dec 2016, 19:30
Another Way:

Question says p*(p+1) = 30X

to check if X is an integer p*(p+1) should have factors 5 and 3*2

Statement 1: says P has factor 5. The number is in multiples of 5. It means that number P ends in 5 or 10.
So its consecutive numbers can be in the multiples of 5,6 or 10,11. Clearly this is not sufficient.

Statement 2: q^2/63 means q has factors 3*7. Clearly not sufficient alone.

Combining these two we resolve to following choice: 5,6 or 15,16... All of them satisfies. Hence C.
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Re: If p and q are two consecutive positive integers and pq= 30x  [#permalink]

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New post 13 Dec 2016, 21:21
subirh wrote:
If p and q are two consecutive positive integers and pq= 30x , is x an integer?

(1) p^2 is divisible by 25.
(2) 63 is a factor of q^2



The question actually asks whether \(pq\) is divisible by \(30=2\times 3 \times 5\) or not.

Note that between two consecutive positive integers, there is always one even number. Hence, \(pq\) is divisible by 2.

(1) \(p^2\) is divisible by \(25=5^2\), so \(p\) is divisible by 5.

If \(p=5 \implies q=6\) is divisible by 3. Hence \(pq\) is divisible by 30.
If \(p=10 \implies q=11\) is not divisible by 3. Hence \(pq\) is not divisible by 30.
Insufficient.

(2) \(q^2\) is divisible by \(63=3^2 \times 7\) so \(q\) is divisible by 3.

If \(q=3 \implies p=2\) is not divisible by 5. Hence \(pq\) is not divisible by 30.
If \(q=6 \implies p=5\) is not divisible by 5. Hence \(pq\) is divisible by 30.
Insufficient.

Combine (1) and (2)
From (1) we have \(p\) is divisible by 5.
From (2) we have \(q\) is divisible by 3.
Hence, \(pq\) will be divisible by \(2 \times 3 \times 5 =30\)

The answer is C.
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Re: If p and q are two consecutive positive integers and pq= 30x  [#permalink]

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New post 07 Sep 2017, 04:53
pq = 30 x
is x an integer?
p,q consecutive

1) p^2/ 25 = 625/25 (example) --> 25 *24 or 25*26 if PQ is considered.
so pq = 30X
30 = 2*5*3 --> 30X
5^2 * 13*2 (for 26) or 5^2 * 3*2^3 = 2*5*3 *x. Therefore, x = 2*5*3 or x = 13*2*5*3 Thus X has two values so insuff. But either way we know that it is likely to be an integer
2) 63 is a factor of q square
--> 63 into a square value. Which is 63*7 = 441 which 21^2.
Either way we don't know about P
1) + 2) = 25 (we know this is in P) * 21 * 21*(q) = 5*3*2*x. Thus, x is an integer.
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If p and q are two consecutive positive integers and pq= 30x  [#permalink]

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New post 17 Sep 2017, 13:01
broall wrote:
subirh wrote:
If p and q are two consecutive positive integers and pq= 30x , is x an integer?

(1) p^2 is divisible by 25.
(2) 63 is a factor of q^2



The question actually asks whether \(pq\) is divisible by \(30=2\times 3 \times 5\) or not.

Note that between two consecutive positive integers, there is always one even number. Hence, \(pq\) is divisible by 2.

(1) \(p^2\) is divisible by \(25=5^2\), so \(p\) is divisible by 5.

If \(p=5 \implies q=6\) is divisible by 3. Hence \(pq\) is divisible by 30.
If \(p=10 \implies q=11\) is not divisible by 3. Hence \(pq\) is not divisible by 30.
Insufficient.

(2) \(q^2\) is divisible by \(63=3^2 \times 7\) so \(q\) is divisible by 3.

If \(q=3 \implies p=2\) is not divisible by 5. Hence \(pq\) is not divisible by 30.
If \(q=6 \implies p=5\) is not divisible by 5. Hence \(pq\) is divisible by 30.
Insufficient.

Combine (1) and (2)
From (1) we have \(p\) is divisible by 5.
From (2) we have \(q\) is divisible by 3.
Hence, \(pq\) will be divisible by \(2 \times 3 \times 5 =30\)

The answer is C.


hi

p is divisible by 5.
q is divisible by 3
so far so good ..

but you conclude, pq is divisible by 2 x 3 x 5....how ..?

where did the "2" come from ....?

please say to me ...

thanks in advance ...
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Re: If p and q are two consecutive positive integers and pq= 30x  [#permalink]

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New post 17 Sep 2017, 13:06
umg wrote:
subirh wrote:
If p and q are two consecutive positive integers and pq= 30x , is x an integer?

(1) p^2 is divisible by 25.
(2) 63 is a factor of q^2

This question is marked as 95% Hard. So, needless to say that this is a Level 750+ question and hence there is no fastest way to solve it. Anyways, following solution is, in one way or other, what everyone above said but I am going to explain it in best way possible.

Universally Available Statements:

i) p and q are two consecutive positive integers
i. a) p > = 1; q > = 1
i. b) p - q = 1 (We do not know which one is bigger and it does not matter)

ii) p*q = 30*x = 2 * 3 * 5 * x.

Question: Is x an Integer?

How to Go about this question: If, Integer x Integer = Integer

How can x be a Non-Integer?
Possible values of x = 1, 2, 3, \(\frac{1}{2}, \frac{2}{3},\frac{3}{5}, \frac{3}{7}, \frac{7}{30}\) and so on such that given conditions are satisfied.

Solution: If 2, 3 & 5 are factors of p, q, or both, then x would always be an Integer. Hence, the question boils down to determining whether p or q are multiples of 2, 3, and 5.

=> p & q are Consecutive Integers. Therefore, their multiple MUST BE EVEN. Hence, either of p or q is a Multiple of 2.

Now we need to determine whether the given statements are sufficient to determine whether p or q are multiples of 3 & 5.

Statement 1: \(p^2\)is divisible by 25

=> p is a Multiple of 5 - Insufficient because we don't know whether 3 is a factor of p or q.

Statement 2: 63 is a factor of q^2

=> q = 3*\(\sqrt{7}\)*z (z is some variable which is a factor of x)

Since q is an Integer,

q = 21*z - Insufficient because we don't know whether 5 is a factor of p or q.
=> q is a Multiple of 3 & 7 - Insufficient because we don't know whether 5 is a factor of p or q.

Statement 1& 2 together:

p is a Multiple of 5 And q is a Multiple of 3 & 7.

=> p*q = Multiple of 2, 3, 5 and 7.
Hence, x will always be an Integer.

Option C - Sufficient.



hi

p is a multiple of 5.
q is a multiple of 3 and 7
so far so good ..

but you conclude, pq is a multiple of 2 x 3 x 5 x 7....how ..?

where did the "2" come from ....?

please say to me ...

thanks in advance ...
Re: If p and q are two consecutive positive integers and pq= 30x &nbs [#permalink] 17 Sep 2017, 13:06
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