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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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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

Answer is D
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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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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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Celestial09
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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I took P=12 for a sample, and ended up picking 12^2 :(
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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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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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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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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 \(p^3 = 6^3\)
Answer D
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One pitfall i was continuously making in these types of questions was that the erroneous assumption that the numerator contained the EXACT amount of the factors of the divisor.

so p can have several factors of 3 as long as there is at least one 2 present

p^2 would be 2^2 x 3^2

which leads us to p^3 and thus the largest divisor!
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sunita123
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


p^2/12 = integer

that means p^2 must at least be 2^2 and 3 that makes p^2 = 36

216 = 6^3 = 2^3 * 3^3

2^3 is possible but not the largest.
2^6 is not possible.
3^3 is possible but not the largest.
6^3 is the largest.

12^2 = (2^2 * 3)^2 = 2^4 * 3^2 which is not possible.

Answer choice D
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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.

Final Answer:
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Rich

If I am understanding this correctly, it sounds like you are saying that any factors of p must have the same ratio of twos to ones (1:1). Is that why E does not work in this situation?

This question is similar: https://gmatclub.com/forum/if-m-is-a-po ... fl=similar

I understand that 16 is not an answer because it is missing a factor of 3. The correct answer of 12 has factors of 2x2x3. The ratio of 2s and 3s in the number 12 is the same as the ratio of 2s and 3s in the number 48 (that the ratio of twos to threes is 2:1). Am I understanding this correctly?
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