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

Since \(3 < x < 100\), then \(1<\frac{x}{3}<33\frac{1}{3}\) (just divide all parts of the inequality by 3).

\(\frac{x}{3}\) should be the square of a prime number, thus \(\frac{x}{3}\) could be 2^2=4, 3^2=9, or 5^2=25.

Answer: B.
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Setting up the equation as follows:

\(\frac{x}{3}\) should be a square

So, say \(\frac{x}{3} = a^2\)

\(x = 3a^2\)

\(3 < 3 a^2 < 100\)

For a = 1; the condition fails

For a = 2; x = 12

For a = 3; x = 27

For a = 5; x = 75

For a = 7; condition fails

Answer = 3 = B

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 condition
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

x/3 is a square or we can take x/3 as y^2; x = y^2 * 3.
Since x should be between 3 and 100, y^2 can take squares 4, 9 and 25.
Answer B.

For x/3 to be a square of a prime number, x = (prime no)^2 x 3

List squares of some prime nos:
4,9,25,49

x can be:

4x3, 9x3 and 25x3

Anything above 25x3 will make x>100, but we are given that 3<x<100

Hence, 3 nos, Option B.

I think Bunnuel has given a perfect explanation that needs no further detailing.

I just want to highlight a point in these types of questions. For the questions that ask "how many values of x".. they will usually be of consecutive numbers. But when the numbers asked are primes, then it is manual counting that is required because prime numbers do not follow any standard pattern. They will definitely be countable and the total will usually be less than 25 in GMAT.

I got confused in this question...why are we considering values 2 and 3 when it is given that 3<x<100...

The question is: for how many values of x is x/3 the square of a prime number.

The answer is: for three values of x (namely for 12, 27, and 75) x/3 is the square of a prime number.

Hope it's clear.
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Hi Bunuel,

Can you perhaps recommend a few similar problems?

Thanks!

Check this one: for-how-many-values-of-k-is-12-12-the-least-common-multiple-86737.html
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What the question stem asks us is how many numbers within 100 are 3 times the square of a prime number?

(2^2)* 3 = 12 - less than 100 and greater than 3
(3^3)* 3 = 27 - less than 100
(5^5)* 3 = 75 - less than 100

We don't have to consider next prime 7 since (7*7)*3 > 100

Thus only 3 possible numbers.

Hi nahid78,

If your question asks how many values of x, then the answer will be 5
The values would be 4, 9, 25, 49, 81

But if you have mistyped the question and we are required to find the values of (x/3)
then the values would be 12, 27, 75. Three values

Attached is a visual that should help.

An easy way to solve this problem is to first 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 . Now solve for x.

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, the product will remain under 100: 3(4) = 12, 3(9) = 27 and 3(25) = 75. Thus, the answer is that 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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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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First, we need to find the range of values for \(\frac{x}{3}\).

Given that \(3<x<100,\) we can divide all sides of the inequality and have

\(\frac{3}{3}<\frac{x}{3}<\frac{100}{3}\)
\(1<\frac{x}{3}<33\frac{1}{3}\)

With this range, there are only a few numbers that are squares of a prime number. They are:
\(2^2=4\), \(3^2=9\), \(5^2=25\)

The final answer is .

Video solution from Quant Reasoning:
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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:

GMAT Assassins aren’t born, they’re made,
Rich

I was plugging in numbers from 1 to 32 then

1.) dividing these numbers by 3 to see if they were prime and
2.) then squaring the primes to see if they fell before the 33 range.

6/3=2 which is a prime and 2^2=4 -> below 33
9/3=3 which is a prime and 3^2=9 --> below 33
15/3=5 which is a prime and 5^2=-->below 33

I still arrive at the correct answer, but it looks like my approach is off based on this form?