T is the set of all positive integers x such that x^2 is a multiple of

Question

T is the set of all positive integers x such that x^2 is a multiple of both 27 and 375. Which of the following integers must be a divisor of every integer x in T?

I. 9
II. 15
III. 27

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

Bunuel · Accepted Answer

OFFICIAL SOLUTION:

The least common multiple of  27 = 3^3  and  375 = 3*5^3  is  3^3*5^3 .

For the integer  x , the minimum value of  x^2  that can be a multiple of  3^3*5^3  is  3^4*5^4 . Hence, the smallest possible value for  x  is  3^2*5^2 . Therefore:
 
 T = \{3^2*5^2, 2*(3^2*5^2), 3*(3^2*5^2), 4*(3^2*5^2), ...\} .
 
The question asks which of the options  must  be a factor of  every  integer  x  in T. Only options I and II meet this criterion because  27 = 3^3 , option III, is not a factor of  3^2*5^2 .

Answer: C

KrishnakumarKA1 · Answer

27 = 3^3
375 = 3 x 5^3
LCM = 3^3 x 5^3

if x^2 is divisible by LCM, then x^2 should have 3^4 x 5^4
therefore x should have 3^2 x 5^2 = 225. therefore every integer in T will be divisible by 225 or factors of 225 which is 1, 3,5,9,15, 25,45, 75, 225.

looking at the option we can see that Option C is correct

NavenRk · Answer

x = sq rt 27*375 = 3^2*5*sqrt5

I 3^2 - yes, its a factor
II 3*5 - yes, its a factor
III 3^3 - not a factor

Hence C

D3N0 · Answer

Ans: C
Solution:
T->{x} where x^2 is a multiple of both 27 and 375 means 3^3 and (5^3)*3
means x must contain 3^2 and 5^2
so with these conditions we know that 9=3^2 and 15=3*5 both have required factors for the divisibility of lowest int for x which is 9*25
but 27 is not a divisor because it can't divide 9*25 fully.

so Ans : C

ScottTargetTestPrep · Answer

Since x^2 is a multiple of both 27 and 375, x^2 is a multiple of the LCM of 27 and 375.

27 = 3^3

375 = 25 x 15 = 5^3 x 3^1

Thus, the LCM of 27 and 375 is 3^3 x 5^3 = 15^3 and x^2 must be a multiple of 15^3. Since x^2 is a perfect square and a multiple of 15^3, the minimum value of x^2 must be be 15 * 15^3 = 15^4. Since x is the square root of x^2, taking the square root of 15^4, the minimum value of x must be 15^2 = 225. We see that both 9 and 15 are factors of 225, but 27 is not.

Answer: C

Kinshook · Answer

Given: T is the set of all positive integers x such that x^2 is a multiple of both 27 and 375. 
Asked: Which of the following integers must be a divisor of every integer x in T?

27 = 3^3
375 = 3*5^3

x^2 = 3^4*5^4*k^2
x = 3^2*5^2*k = 9*25*k = 175k

I. 9: YES
II. 15 = 3*5: YES
III. 27 = 3^3: NO

IMO C

Ffsimlildig · Answer

Hi  Bunuel , can you please explain this sentence further

For an integer x, the least value of x^2 which is a multiple of 3^3*5^3 is 3^4*5^4, so the least value of x is 3^2*5^2.

why must the least value of x^2 be 3^4*5^4?

thanks

Bunuel · Answer

Since x is an integer, x^2 is the square of an integer, thus the powers of its primes must be even. For instance, if  x = p^a*q^b*... , then  x^2 = p^{2a}*q^{2b}*... . Thus, the least value of  x^2  such that it is a multiple of  3^3*5^3  is  3^4*5^4 .

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

T is the set of all positive integers x such that x^2 is a multiple of

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

T is the set of all positive integers x such that x^2 is a multiple of

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards