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x ^2 is divisible by both 40 and 75. if x has exactly 3

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x ^2 is divisible by both 40 and 75. if x has exactly 3 [#permalink] New post 05 Mar 2016, 11:35
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21% (00:55) correct 78% (01:42) wrong based on 41 sessions
x² is divisible by both 40 and 75. if x has exactly 3 distinct prime factors, which of the following could be the value of x?
Indicate all values that apply.

A) 30
B) 60
C) 200
D) 240
E) 420
[Reveal] Spoiler: OA
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Re: x ^2 is divisible by both 40 and 75. if x has exactly 3 [#permalink] New post 12 Mar 2016, 12:57
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Explanation


The two clues given in the question are the following:
  • \(x ^2\) is divisible by both 40 and 75. Lets rewrite 40 as \(2^3*5\) and 75 as \(5^2*3\)
  • x has exactly 3 distinct prime factors
Now lets play a game of elimination based on these:
  • \(30 = 2*3*5\), It satisfies the second clue however it fails satisfying the first one since x^2 or 900is not divisible by 40.
  • \(60 = 2^2*3*5\), It satisfies the second clue and also \(60^2 = 2^4*3^2*5^2\) so \(x^2\) is also divisible by both 40 and 50(check the factors).
  • \(200 = 2^3*5^2\), It doesn't satisfy the second clue since it has only two distinct prime factors.
  • \(240 = 2^4*3*5\), It satisfies the second clue and also \(240^2 = 2^8*3^2*5^2\) so \(x^2\) is also divisible by both 40 and 50(check the factors).
  • \(420 = 2^2*3*5*7\), It doesn't satisfy the second clue since it has four distinct prime factors.

So the the answers are B & D.
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Re: x ^2 is divisible by both 40 and 75. if x has exactly 3 [#permalink] New post 25 Jun 2016, 15:05
Hi, Thanks for your respond. It was very helpful. I understand how to solve this question looking at the prime factorization and seeing that Answer choice E. 420 has an extra prime factorization of 7. However, Can you please explain further why this is a wrong answer choice? For example
420^2 = 176400
176400 / 45 = 3920
176400 / 75 = 2352.

As you can see 420^2 is divisible by both 40 and 75 and as a result, I chose that as the correct answer? Can you please explain why my logic is incorrect?

THANKS SO MUCH AS USUAL!!!
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Re: x ^2 is divisible by both 40 and 75. if x has exactly 3 [#permalink] New post 25 Jun 2016, 15:21
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Hi,

420 passes the divisibility test but it has one prime factor i.e. "7" more than the given allowed number of prime factors which is 3. So the prime factors allowed for possible candidate solution are 2, 3 and 5. Hence B and D.

A candidate solution must satisfy both divisibility criterion and number of prime factors allowed.


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Re: x ^2 is divisible by both 40 and 75. if x has exactly 3 [#permalink] New post 01 Aug 2020, 07:25
Expert's post
greMS15 wrote:
x² is divisible by both 40 and 75. if x has exactly 3 distinct prime factors, which of the following could be the value of x?
Indicate all values that apply.

A) 30
B) 60
C) 200
D) 240
E) 420


-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------


x² is divisible by 40
40 = (2)(2)(2)(5)
So, x² = (2)(2)(2)(5)(?)(?)(?)...
Note: the (?)'s represent additional prime factors that COULD be in the prime factorization of x²

Question: If x² = (2)(2)(2)(5)(?)(?)(?)..., what can we say about x?
Well, if we were to take the product (2)(2)(2)(5)(?)(?)(?) and break it into two EQUIVALENT products (x and x), we can conclude that each individual product must have at least TWO 2's and ONE 5.
In other words, the prime factorization of x must have least TWO 2's and ONE 5
We can write: x = (2)(2)(5)(?)(?)...

x² is divisible by 75
75 = (3)(5)(5)
So, x² = (3)(5)(5)(?)(?)(?)...
So, the prime factorization of x must have least ONE 3 and ONE 5
We can write: x = (3)(5)(?)(?)...

When we COMBINE both pieces of information, we know that: the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5
In other words: x = (2)(2)(3)(5)(?)(?)...


If x has exactly 3 distinct prime factors, which of the following could be the value of x?
This tells us that 2, 3 and 5 are the ONLY prime factors of x.

Now let's examine the answer choices....
A) 30 = (2)(3)(5)
This BREAKS the condition that the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5
So, x cannot equal 30

B) 60 = (2)(2)(3)(5)
60 works perfectly!
It satisfies the condition that the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5, and it satisfies the condition that 2, 3 and 5 are the ONLY prime factors of x

C) 200 = (2)(2)(2)(5)(5)
This BREAKS the condition that the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5
So, x cannot equal 200

D) 240 = (2)(2)(2)(2)(3)(5)
240 works perfectly!
It satisfies the condition that the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5, and it satisfies the condition that 2, 3 and 5 are the ONLY prime factors of x

E) 420 = (2)(2)(3)(5)(7)
This satisfies the condition that the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5, BUT it BREAKS the condition that 2, 3 and 5 are the ONLY prime factors of x

Answer: B and D

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.
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Re: x ^2 is divisible by both 40 and 75. if x has exactly 3   [#permalink] 01 Aug 2020, 07:25
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