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# If P and Q represent the hundreds and tens digits, ..x=8PQ2.

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Senior Manager
Joined: 21 Oct 2013
Posts: 416
If P and Q represent the hundreds and tens digits, ..x=8PQ2.  [#permalink]

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23 Jun 2014, 09:49
8
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Difficulty:

65% (hard)

Question Stats:

59% (01:57) correct 41% (01:26) wrong based on 136 sessions

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If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

(1) P=4
(2) Q=0

OE
According to Stat. (2), 8P02 will always end with an 02. Remember that For a number to be divisible by 4, the number's last two digits must form a number that is divisible by 4. Since 02 is not a number divisible by 4, the entire number, 8P02 is not divisible by 4. A number that is indivisible by 4 cannot be divisible by 8=4*2, so stat. (2) tells you that the answer to the question stem is a definite "no" - which is sufficient.

If you're not sure, plug in the different digits (0-9) for P and see for yourself how none of the plug-ins (8102, 8202, 8302, etc.) yields a number that is divisible by 8.
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Joined: 02 Sep 2009
Posts: 55271
If P and Q represent the hundreds and tens digits, ..x=8PQ2.  [#permalink]

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23 Jun 2014, 09:57
4
3
goodyear2013 wrote:
If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

(1) P=4
(2) Q=0

OE
According to Stat. (2), 8P02 will always end with an 02. Remember that For a number to be divisible by 4, the number's last two digits must form a number that is divisible by 4. Since 02 is not a number divisible by 4, the entire number, 8P02 is not divisible by 4. A number that is indivisible by 4 cannot be divisible by 8=4*2, so stat. (2) tells you that the answer to the question stem is a definite "no" - which is sufficient.

If you're not sure, plug in the different digits (0-9) for P and see for yourself how none of the plug-ins (8102, 8202, 8302, etc.) yields a number that is divisible by 8.

If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

For a number to be divisible by 2 the last digit must be divisible by 2 (so the last digit must be even);
For a number to be divisible by 4 the last two digits must be divisible by 4 (04, 08, 12, 16, ..., 96);
For a number to be divisible by 8 the last three digits must be divisible by 8 (008, 012, 016, ..., );
etc.

(1) P=4. If Q=0, then the answer is NO, because 402 is not divisible by 8, but if Q=3, then the answer is YES, because 432 is divisible by 8. Not sufficient.

(2) Q=0. In this case the last tow digits are 02, so x is not divisible by 4, which means that it's not divisible by 8 either. Sufficient.

Hope it's clear.
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Re: If P and Q represent the hundreds and tens digits, ..x=8PQ2.  [#permalink]

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24 Jul 2014, 05:31
Bunuel why we are checking for 4 ... why don't the we check directly for divisibility by 8 .. for last three digits
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Re: If P and Q represent the hundreds and tens digits, ..x=8PQ2.  [#permalink]

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24 Jul 2014, 05:46
GmatDestroyer2013 wrote:
Bunuel why we are checking for 4 ... why don't the we check directly for divisibility by 8 .. for last three digits

Which part are you referring to? The second statement?

From (2) we can get that the number cannot be divisible by 4, so no need to check for 8, because it's clear that since it's not divisible by 4 then it also won't be divisible by 8.
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Re: If P and Q represent the hundreds and tens digits,  [#permalink]

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07 Aug 2014, 06:39
goodyear2013 wrote:
If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

(1) P=4
(2) Q=0

Here's how I solved this question:

I started with statement 2 because it seemed easier to tackle a 0 than a 4.

1) Assuming statement 2, I have to answer: is 8P02 divisible by 8?
2) In doing long division, the first 8 is just a distraction because it will never leave a remainder for the hundreds digit. So I started by just removing it entirely.
3) Now I end up with P02 divisible by 8?
4) I saw that the only multiple of 8 that ends in a "2" is 32. Therefore I would need a "P0" value that leaves a remainder of 3. Since 8 is an even number and so is 0, I know that there is no way for there to be a remainder of 3. Therefore I can say statement 2 is sufficient to answer the question stem: no, X is not divisible by 8. NOTE: This can be tricky, because some people will get to this point and think that because the answer to the question stem is "no" that statement two is not correct. Therein lies the trickery of the data sufficiency.

Here, I can eliminate options A, C, E.

Now I need to test if statement 1 is sufficient.

1) Assuming statement 1, I have to answer: is 84Q2 divisible by 8?
2) Again I can just exclude the 8 to make my calculations simpler.
3) Now I end up with is 4Q2 divisible by 8?
4) Next, I just plugged in a value to test it. Since I know that a value of 0 definitely makes the number NOT divisible by 8 (discovered from my solving of statement 2), I try to look for a way to make the number divisible by 8.
5) As I did my work for statement 2, I had discovered that I would need there to be a remainder of 3 in the 10s digit to make the number divisible by 8. Well, it looks like I can do that by making Q=3. Therefore, I can say that statement 1 is not sufficient.

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If P and Q represent the hundreds and tens digits,  [#permalink]

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07 Aug 2014, 21:51
goodyear2013 wrote:
If P and Q represent the hundreds and tens digits, respectively, in the four-digit number x=8PQ2, is x divisible by 8?

(1) P=4
(2) Q=0

(1) P = 4
X = 8000 + 100P+ 10Q + 2 = 8000+ 400 + 10Q + 2
Since 8400 is divisible by 8, (10Q +2) has to be divisible by 8. Q is from 1 to 9. If you check, there are 32 and 72 that satisfy the requirement. So 8432 is divisible by 8 but 8452 is not divisible by 8.

(2) Q = 0
X= 8000 + 100P + 2
Same reasoning, (100P+2 ) has to be divisible by 8. P is from 1 to 9.
We have, 102, 202,....902. For the numbers to be divisible by 8, they have to be divisible by 4, meaning the last two digits have to be divisible by 4. The last two digits are 02, not divisible by 4. So 8PQ2 is not divisible by 8 --> SUFFICIENT.

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Re: If P and Q represent the hundreds and tens digits,  [#permalink]

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08 Aug 2014, 00:41
Criterions of divisibility by 2,4, and 8:
2: the last digit of a number is divisible by 2;
4: the last two digits of a number form the integer divisible by 4
8: the last three digits of a number form the integer divisible by 8

In the problem $$x$$ is divisible by 2, but we don't know about divisibility by 8.

(1) If $$P=4$$, we have number $$x=84Q2$$, which is divisible by 8 if $$4Q2$$ is divisible by 8. The answer depends on $$Q$$: if $$Q$$ is 0, $$x$$ is not divisible by 8, but if $$Q$$ is 3, $$x$$ is divisible by 8. Insufficient

(2) If $$Q=0$$, we have number $$x=8P02$$, which is not divisible by 4, since $$02$$ is not divisible by 4. Hence, $$x$$ is not divisible by 8. Sufficient

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Re: If P and Q represent the hundreds and tens digits, ..x=8PQ2.  [#permalink]

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21 Mar 2019, 17:01
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Re: If P and Q represent the hundreds and tens digits, ..x=8PQ2.   [#permalink] 21 Mar 2019, 17:01
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