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Re: If 3 < x < 100, for how many values of x is x/3 the square
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15 Jun 2016, 05:04

2

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.

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

Re: If 3 < x < 100, for how many values of x is x/3 the square
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30 Aug 2017, 13:45

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Top Contributor

1

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:

Re: If 3 < x < 100, for how many values of x is x/3 the square
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02 Sep 2017, 06:18

2

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.

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

Square of a prime number means that the resulting square has prime number as its root.

For e.g. 9 has 3 as its root so 9 is square of a prime number because 3 is prime But 16 is not a square of a prime number because it’s root is 4 which is not prime

Re: If 3 < x < 100, for how many values of x is x/3 the square
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25 Apr 2019, 08:09

Have to read more carefully... wasted almost a minute thinking the question said the the prime number must be >3 rather than x (which would mean only 5 fits).

An easy way to do this is to revert the equation... x = prime#² * 3 Since x < 100, we can just test 2²*3 = 12 3²*3 = 27 5²*3 = 75 7²* 3 = 147, too big.