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If p is a positive integer and p^2 is divisible by 12
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29 Dec 2014, 16:51
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66% (01:37) correct 34% (01:55) wrong based on 281 sessions
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If p is a positive integer and p^2 is divisible by 12, then the largest positive integer that must divide p ^3 is (A) 2^3 (B) 2^6 (C) 3^3 (D) 6^3 (E) 12^2
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Re: If p is a positive integer and p^2 is divisible by 12
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30 Dec 2014, 21:28
Hi sunita123, This question is essentially about prime factorization and how exponents "work." When a question uses an exponent, you have to remember what the exponent implies: Here, we have P^2, which means (P)(P) For example, if P is a positive integer and P^2 = 100, then P = 10 This can be written in a variety of ways: P^2 = 100 P^2 = 10^2 P^2 = (10)(10) P^2 = [2x5][2x5] The point of all of this is that each "P" has to be made up of the same "pieces"  in this example, P is made up of one "2" and one "5." Here, we're told that P is a positive integer and P^2 is DIVISIBLE by 12. This causes a bit of extra work for us because 12 = (2)(2)(3). P CANNOT = 12 though because there aren't enough "pieces" to make both Ps the same. We can put a "2" into each "P" but we can't split up the 3.... P^2 = (P)(P) = (2 x something)(2 x something) ....but the 3 has to go in there somewhere.... If we had ANOTHER 3 though, then we'd have P^2 = (2)(2)(3)(3) and then we can put one "2" and one "3" into each P... P^2 = (2x3)(2x3) = 36 P = 6 Having that extra 3 is the simplest/smallest value for P that fits this question. From here, P^3 = 6^3. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If p is a positive integer and p^2 is divisible by 12
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29 Dec 2014, 19:56
Since p is an integer so p cannot have a 2 and sqrt 3 (because squaring this will give us a 2^2 and 3 (making the product as 12, and making p^2 as a multiple of 12)) p^2 is divisible by 12 (12 = 2*2*3), So, p should have at least one 2 and one 3 so that p^2 has a 2^2 and two 3 So, p will have a 2 and a 3. Or p will be a multiple of 6 So, largest possible integer than should divide p^3 is 6^3 So, Answer will be D Hope it helps! sunita123 wrote: If p is a positive integer and p^2 is divisible by 12, then the largest positive integer that must divide p ^3 is
(A) 2^3 (B) 2^6 (C) 3^3 (D) 6^3 (E) 12^2
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Re: If p is a positive integer and p^2 is divisible by 12
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30 Dec 2014, 21:09
Answer = (D) 6^3 Least value of p for which p^2 divisible by 12 is 6 \(\frac{6^2}{12} = 3\) Largest value which can divide 6^3 is 6^3
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Re: If p is a positive integer and p^2 is divisible by 12
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05 May 2015, 19:17
If p^2 is divisible by 12 then we can conclude that the prime box of p has atleast 3 & 2 in it. (note that we are not sure if there is a 4)
So p^3 will be divisible by (3x2)^3, that is 6^3
Answer is D



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Re: If p is a positive integer and p^2 is divisible by 12
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06 May 2015, 03:56
In questions like this that give and ask information about divisibility of higher powers of an integer, it's a good idea to start with the prime factorized form of the integer itself. That is, as the first step, express \(P =\)\(P1^{a}*P2^{b}*P3^{c}*P4^{d}\) . . . where P1, P2, P3, P4 etc. are prime numbers and a, b, c, d etc. are nonnegative integers. This means \(P^2\) =\(P1^{2a}*P2^{2b}*P3^{2c}*P4^{2d}\) . . . We are given that \(P^2\) is divisible by \(12=\) \(2^{2}*3^{1}\) This means, P1 = 2 and 2a >= 2 That is, a > = 1. So, minimum possible value of a = 1 Also, P2 = 3 and 2b > = 1 That is, b > = 0.5 But b must be an integer. So, minimum possible value of b = 1 So, we see that P = \((2^{1}*3^{1})(something)\) . . . We don't have any idea about what the value of this 'something' is, but we can say for sure that P must be divisible by \((2^{1}*3^{1})\) and so, \(P^3\) must be divisible by \((2^{3}*3^{3})\) Hope this was useful! Japinder
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Re: If p is a positive integer and p^2 is divisible by 12
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06 May 2015, 10:41
Hello! Apologies if I am puting an absurd question. What if there was 6^4 and also 6^3 in the answer options. Right now we selected 6^3... but what if with 6^3 there is also 6^4 ? will we choose 6^4? Thanks Celestial



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Re: If p is a positive integer and p^2 is divisible by 12
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06 May 2015, 10:58
Celestial09 wrote: Hello! Apologies if I am puting an absurd question. What if there was 6^4 and also 6^3 in the answer options. Right now we selected 6^3... but what if with 6^3 there is also 6^4 ? will we choose 6^4? Thanks Celestial All we know is that the prime box of p has at least One 3 and One 2. So p^3 will have at least Three 3s' and Three 2s' (3^3 x 2^3) hence the highest factor possible is 6^3. We do not have any evidence to conclude that p^3 can have more than three 3s' and 2s'. Hence 6^4 can't be a factor of p. Please note it is possible that p has more than three 3s' and 2s' but we cant conclude that from the statement. Hope it is clear. Regards, Aj



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Re: If p is a positive integer and p^2 is divisible by 12
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19 Aug 2016, 08:59
I took P=12 for a sample, and ended up picking 12^2



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Re: If p is a positive integer and p^2 is divisible by 12
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19 Aug 2016, 11:34
P^2 has atleast two 2's and two 3's (squares has even powers of integers) P must thus contain atleast one 2 and one 3. P^3=2*2*2*3*3*3*x P^3=6^3*x P^3 is thus divisible by atleast at most 6^3 given the information known to us. Posted from my mobile device
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Re: If p is a positive integer and p^2 is divisible by 12
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21 Sep 2016, 06:46
sunita123 wrote: If p is a positive integer and p^2 is divisible by 12, then the largest positive integer that must divide p ^3 is
(A) 2^3 (B) 2^6 (C) 3^3 (D) 6^3 (E) 12^2 the wording is kind of confusing...just because of rereading everything i lost ~20 sec. since p^2 is divisible by 12, and 12 contains 2^2 and one 3, it must be true that p^2 must contain at least one more 3. since we are given p^3, then 6^3 would work, as 6 has both 2 and 3, and everything is raised to the power of 3. we know for sure that in this case p^3 is divisible by 6^3. 12^2 on the other hand...contains 2^4 * 3^2  we are not sure whether we have another factor of 3... so D works.



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Re: If p is a positive integer and p^2 is divisible by 12
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02 Jan 2018, 02:55
If p*p is divisible by 12, that means each p must consist 2 and 3; then p has to be 6 only. Now, p^3 becomes 6^3 ie. 2^3 * 3^3. Only D is correct option.



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Re: If p is a positive integer and p^2 is divisible by 12, then the larges
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17 Oct 2018, 09:48




Re: If p is a positive integer and p^2 is divisible by 12, then the larges &nbs
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