If 3 < x < 100, for how many values of x is x/3 the sq

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If 3 < x < 100, for how many values of x is x/3 the square of a prime number?

(A) Two
(B) Three
(C) Four
(D) Five
(E) Nine

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by [email protected] » Sat Apr 29, 2017 11:58 pm
rsarashi wrote:If 3 < x < 100, for how many values of x is x/3 the square of a prime number?

(A) Two
(B) Three
(C) Four
(D) Five
(E) Nine

OAB
We have a situation: x/3 the square of a prime number such that 3 < x < 100.

3 < x < 100 => 1 < x/3 < 100/3 => 1 < x/3 < 33.33 => 1 < p^2 < 33.33; where p is a prime number

Let us list down few prime numbers; they are 2, 3, 5, 7, 11, etc.

Since x/3 is a square of a prime number, only prime numbers that would qualify are 2, 3, & 5.

The correct answer: B

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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by [email protected] » Sun Apr 30, 2017 7:22 am
rsarashi wrote:If 3 < x < 100, for how many values of x is x/3 the square of a prime number?

(A) Two
(B) Three
(C) Four
(D) Five
(E) Nine

OAB
We want values of x (where 3 < x < 100) such that x/3 is the square of a prime number.
So, let's start checking squares of prime numbers.
Some prime numbers are 2, 3, 5, 7, 11, etc

2² = 4 and (3)(4) = 12. So, x = 12 meets the given conditions.
3² = 9 and (3)(9) = 27. So, x = 27 meets the given condition12
5² = 25 and (3)(25) = 75. So, x = 75 meets the given conditions.
7² = 49 and (3)(49) = 147. No good. We need values of x such that 3 < x < 100

So, there are exactly 3 values of x that meet the given conditions.
Answer: B

Cheers,
Brent
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by rsarashi » Wed May 03, 2017 9:19 am
2² = 4 and (3)(4) = 12. So, x = 12 meets the given conditions.
3² = 9 and (3)(9) = 27. So, x = 27 meets the given condition12
5² = 25 and (3)(25) = 75. So, x = 75 meets the given conditions.
7² = 49 and (3)(49) = 147. No good. We need values of x such that 3 < x < 100
Hi Brent ,

Thank you so much for your reply.

Just a quick question can you please advise that why did you multiply with 3? We have to divide x/3.

Please explain sir.

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by [email protected] » Wed May 03, 2017 9:27 am
rsarashi wrote:
2² = 4 and (3)(4) = 12. So, x = 12 meets the given conditions.
3² = 9 and (3)(9) = 27. So, x = 27 meets the given condition12
5² = 25 and (3)(25) = 75. So, x = 75 meets the given conditions.
7² = 49 and (3)(49) = 147. No good. We need values of x such that 3 < x < 100
Hi Brent ,

Thank you so much for your reply.

Just a quick question can you please advise that why did you multiply with 3? We have to divide x/3.

Please explain sir.
Great question.

First off, we want x/3 to be the square of a prime number.
Since all prime numbers are integers, the square of a prime number will be an integer.

In order for x/3 to be an integer, it must be the case that x is divisible by 3.
Another way to put it is that x must be a multiple of 3.
This why I took each squared value and multiplied it by 3.

Cheers,
Brent
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by rsarashi » Sat May 06, 2017 6:46 pm
Great question.

First off, we want x/3 to be the square of a prime number.
Since all prime numbers are integers, the square of a prime number will be an integer.

In order for x/3 to be an integer, it must be the case that x is divisible by 3.
Another way to put it is that x must be a multiple of 3.
This why I took each squared value and multiplied it by 3.

Hi Brent ,

Perfect! All clear.

Thanks

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by [email protected] » Tue Jan 02, 2018 11:10 am
rsarashi wrote:If 3 < x < 100, for how many values of x is x/3 the square of a prime number?

(A) Two
(B) Three
(C) Four
(D) Five
(E) Nine
We can write out all the perfect squares below 100 that result from squaring a prime number. The prime numbers to consider are 2, 3, 5, and 7. The next prime number, 11, yields 121 when it is squared, which is too large, and so we only consider the following four squared prime numbers:

4, 9, 25, 49

(Keep in mind that it's useful to have all the perfect squares below 100 memorized and note that 4 = 2^2, 9 = 3^2, 25 = 5^2, and 49 = 7^2.)

Next, we can write the question stem as an equation.

x/3 = (prime)^2

Solving for x, we have:

x = 3(prime)^2

From our list, we see that there are 3 values (4, 9, and 25) that, when we multiply them by 3, have a product that is less than 100: 3(4) = 12, 3(9) = 27, and 3(25) = 75. Thus, there are 3 values (12, 27, and 75) such that x/3 is the square of a prime number.

Answer: B

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Hi All,

We’re told that 3 < X < 100. We’re asked for the number of possible values for X that would make X/3 the SQUARE of a PRIME NUMBER. From the answer choices, we know that there are at least 2 values, but no more than 9 values, that fit what we’re asked for, so we should be able to list them all out without too much difficulty.

To start, it’s worth noting that since X < 100, we know that the value of X/3 will be 33 or less. We’re looking for SQUARES of PRIME NUMBERS, so the number of possibilities is going to be really ‘limited’; we can start at the smallest Prime and work up…

2; 2^2 = 4…. If X=12, then 12/3 = 4 which is a square of a prime

3: 3^2 = 9… If X = 27, then 27/3 = 9 which is a square of a prime

5: 5^2 = 25… If X = 75, then 75/3 = 25 which is a square of a prime

7: 7^2 = 49… Here, we would need X to be GREATER than 100 (specifically 147, which is not allowed).

Thus, there are only 3 possible values of X that fit the ‘restrictions’ in the prompt.

Final Answer: B

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