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# If p is a positive integer and p^2 is divisible by 12

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Manager
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If p is a positive integer and p^2 is divisible by 12  [#permalink]

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29 Dec 2014, 16:51
4
6
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Difficulty:

35% (medium)

Question Stats:

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  [#permalink]

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30 Dec 2014, 21:28
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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.

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Re: If p is a positive integer and p^2 is divisible by 12  [#permalink]

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29 Dec 2014, 19:56
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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

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  [#permalink]

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30 Dec 2014, 21:09
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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  [#permalink]

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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 at-least 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

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Re: If p is a positive integer and p^2 is divisible by 12  [#permalink]

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06 May 2015, 03:56
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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 non-negative 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!

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Re: If p is a positive integer and p^2 is divisible by 12  [#permalink]

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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
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Re: If p is a positive integer and p^2 is divisible by 12  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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.

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Re: If p is a positive integer and p^2 is divisible by 12  [#permalink]

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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 re-reading 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  [#permalink]

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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  [#permalink]

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17 Oct 2018, 09:48
Since, $$p^2$$ is divisible by 12 , $$2^2*3$$
hence the least value of p = 2*3= 6
the largest integer which must divides $$p^3 = 6^3$$
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Re: If p is a positive integer and p^2 is divisible by 12, then the larges &nbs [#permalink] 17 Oct 2018, 09:48
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