If p is a postive integer less than 29 and q is the remainder when 29 is divided by p, what is the value of q?
(1) p is a two-digit number.
(2) p = 3^k, where k is a positive integer.
C
however, i believe statement 2 is sufficient to answer the question. if k=1 then the remainder is 2---> 29/3 remainder is 2
if k = 2 then p is 9 and still the remiander is 2
if k = 3 then p is 27 and still the remainder is 2. the question stem says that p is less than 29 so it limits the property of p to 3, 9 or 27. in all three cases, the remiander is 2.
jeff sackmann total gmat math question
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- hemant_rajput
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you are correct.sana.noor wrote:If p is a postive integer less than 29 and q is the remainder when 29 is divided by p, what is the value of q?
(1) p is a two-digit number.
(2) p = 3^k, where k is a positive integer.
C
however, i believe statement 2 is sufficient to answer the question. if k=1 then the remainder is 2---> 29/3 remainder is 2
if k = 2 then p is 9 and still the remiander is 2
if k = 3 then p is 27 and still the remainder is 2. the question stem says that p is less than 29 so it limits the property of p to 3, 9 or 27. in all three cases, the remiander is 2.
I personally find the jeff sackman's OA can't be taken for granted all the time.
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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If p is a postive integer less than 29 and q is the remainder when 29 is divided by p, what is the value of q?
(1) p is a two-digit number.
(2) p = 3^k, where k is a positive integer.
however, i believe statement 2 is sufficient to answer the question. if k=1 then the remainder is 2---> 29/3 remainder is 2
if k = 2 then p is 9 and still the remiander is 2
if k = 3 then p is 27 and still the remainder is 2. the question stem says that p is less than 29 so it limits the property of p to 3, 9 or 27. in all three cases, the remiander is 2.
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No just Statement 2 is not sufficient. Clearly p==q. Now if you just consider B than their can be multiple values of Q.3,9 or 27. If you include 1 than only 1 value remains(27).
(1) p is a two-digit number.
(2) p = 3^k, where k is a positive integer.
however, i believe statement 2 is sufficient to answer the question. if k=1 then the remainder is 2---> 29/3 remainder is 2
if k = 2 then p is 9 and still the remiander is 2
if k = 3 then p is 27 and still the remainder is 2. the question stem says that p is less than 29 so it limits the property of p to 3, 9 or 27. in all three cases, the remiander is 2.
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No just Statement 2 is not sufficient. Clearly p==q. Now if you just consider B than their can be multiple values of Q.3,9 or 27. If you include 1 than only 1 value remains(27).
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I received a message about this thread.
there's definitely a misunderstanding here, because you wrote "clearly p=q" (unless 2 equals signs means something else), but it's actually impossible for p and q to be the same here.
If you divide something by p, then the remainder (= q) will always be less than p, That's how remainders work.
yes.however, i believe statement 2 is sufficient to answer the question. if k=1 then the remainder is 2---> 29/3 remainder is 2
if k = 2 then p is 9 and still the remiander is 2
if k = 3 then p is 27 and still the remainder is 2. the question stem says that p is less than 29 so it limits the property of p to 3, 9 or 27. in all three cases, the remiander is 2.
no. those are values of p, not values of q.No just Statement 2 is not sufficient. Clearly p==q. Now if you just consider B than their can be multiple values of Q.3,9 or 27. If you include 1 than only 1 value remains(27).
there's definitely a misunderstanding here, because you wrote "clearly p=q" (unless 2 equals signs means something else), but it's actually impossible for p and q to be the same here.
If you divide something by p, then the remainder (= q) will always be less than p, That's how remainders work.
Ron has been teaching various standardized tests for 20 years.
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Thanks Ron. I think sometimes i read the question too fast in order to gain time.This is one of those times.lunarpower wrote:I received a message about this thread.
yes.however, i believe statement 2 is sufficient to answer the question. if k=1 then the remainder is 2---> 29/3 remainder is 2
if k = 2 then p is 9 and still the remiander is 2
if k = 3 then p is 27 and still the remainder is 2. the question stem says that p is less than 29 so it limits the property of p to 3, 9 or 27. in all three cases, the remiander is 2.
no. those are values of p, not values of q.No just Statement 2 is not sufficient. Clearly p==q. Now if you just consider B than their can be multiple values of Q.3,9 or 27. If you include 1 than only 1 value remains(27).
there's definitely a misunderstanding here, because you wrote "clearly p=q" (unless 2 equals signs means something else), but it's actually impossible for p and q to be the same here.
If you divide something by p, then the remainder (= q) will always be less than p, That's how remainders work.
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yeah, you shouldn't do that. if anything, you should read the questions very slowly and carefully.Blue_Skies wrote:Thanks Ron. I think sometimes i read the question too fast in order to gain time.This is one of those times.
even if you do that, you'll be spending a few extra seconds at most.
if "time pressure" is a problem, that doesn't mean you have to work/read faster. instead, it means that you have to QUIT faster when you are STUCK.
if you can honestly get to the point where you quit and move on (either by guessing, or by trying another method on the same problem) at the very moment you truly become stuck, then you will have no time-management problems at all.
Ron has been teaching various standardized tests for 20 years.
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
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