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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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Updated on: 12 Oct 2012, 01:10
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9
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Difficulty:

95% (hard)

Question Stats:

44% (09:01) correct 56% (01:53) wrong based on 213 sessions

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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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subirhait

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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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.

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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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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 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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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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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 $$2 \times 3 \times 5 =30$$

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

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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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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 $$2 \times 3 \times 5 =30$$

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 ....?

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

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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 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 ....?

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