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!


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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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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!

Japinder
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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

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*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.