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

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Re: If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 15 Jun 2016, 05:04
Walkabout 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


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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Re: If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 13 Sep 2016, 22:33
Walkabout 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

Assume x=+int
mistake mine: (x/3)^2=P wrong
(x/3) = P^2

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If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 22 Nov 2016, 00:14
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Great Official Question.
Here is what i did =>
x/2= P^2 where P is a prime number
Hence x=Prime^2*3

NOTE => Its a good idea to remember atleast the first Ten primes. Even Better if you know all the primes upto 100.
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
101



Now putting P=2,3,5 we get value of x in the bound (3,100)
hence Three is our answer.

Hence B
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Re: If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 25 Mar 2017, 11:09
since 3<x<100
then 1<x/3<33,3
then those numbers are 2, 3, 5
Hence 3 numbers
Answer is B

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If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 25 Mar 2017, 23:28
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

The question confusingly adds x/3... lol so the square is the value of x/3

so we take first prime numbers 2,3,5,7,11 and are 4,9,25,49,121..


Since these values should be x/3 and the value of x will be 4*3,9*3,25*3,49*3..... which are 12,27,75,147....


we only need values below 100 so we have only 12,27 and 75 Three is the answer.B

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Re: If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 15 Apr 2017, 13:20
Bunuel wrote:
Walkabout 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


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.


How do you know to divide the inequality by 3. is there a general strategy for this?

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If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 15 Apr 2017, 16:37
Walkabout 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


let x/3=y^2
3y^2<100
y^2<34
y^2 can only=4, 9, or 25
3
B

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Re: If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 30 Aug 2017, 13:45
Expert's post
Top Contributor
Walkabout 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 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:
[Reveal] Spoiler:
B


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Re: If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 02 Sep 2017, 06:18
Walkabout 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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Re: If 3 < x < 100, for how many values of x is x/3 the square [#permalink]

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New post 31 Oct 2017, 14:33
I did an excel spreadsheet to understand this. However, the explanation from other people to divide 100/3 = 33.33... helps to reduce the amount of work.

thank you all!
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Re: If 3 < x < 100, for how many values of x is x/3 the square   [#permalink] 31 Oct 2017, 14:33

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