If 144/x is an integer and 108/x is an integer, which of the following

This is a number properties question.

There are two approaches you can take.

Proof approach:
For 144/x to be an integer X can be any multiple of the following 2,2,2,2,3,3
For 108/x to be an integer x can be any multiple of the following 2,2,3,3,3
The largest possible value for X will be the multiple of the common elements 2x2x3x3 = 36
If you know your number props well this is quick and easy

Back solve approach:
Looking at the answer options
Statement I) 9/x is an integer – this is true if x<=9 and a factor of 9. So this is true when x is 3 or 9. But X could be 12 and still meet conditions. Not True.
Statement II) 12/x is an integer – this is true if x<=12 and a factor of 12. So this is true when x is 2,3,4,12. But X could be 18 and still meet conditions. Not True.
We could go and solve this (and if you do you’ll see that only 36 meets all criteria) but from the answer choices we can see that the only answer choice that does not include 1 and 2 is choice B.

I prefer the proof approach, its neater and quick if you know your number props.

from the proof approach, the smallest possible value for x is 2 and the largest is 36.

So basically we just need to test these 2 extremes. Which shows 36 only will work.

36 is the greatest common factor for 144 and 108, so the greatest possible value of X = 36

so, only III is true for all values X

My answer is B

Hi,

I didnt get this one. The q askes for which one of the following is an integer. So i plugged in nos. to see, and found that each option is has divisors. On what basis is the answer 36 as d qustn askes for "is an integer" and not greatest factor?

Pls explain,
Thnx

If 144/x is an integer and 108/x is an integer, which of the following must be true?

I. 9/x is an integer
II. 12/x is an integer
III. 36/x is an integer

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

The question asks which of the following MUST be true, not COULD be true. The largest possible value of x is 36, GCD of 144 and 108, and if x=36 then ONLY III is true.

Answer: B.

Check more Must or Could be True Questions to practice: search.php?search_id=tag&tag_id=193

Hope it helps.
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Why is statement-2 to not true?

Statement II) 12/x is an integer –
this is true if x<=12 and a factor of 12. So this is true when x is 2,3,4,12. Is this not sufficient as both 144 and 108 are divisible by any of the numbers 2,3,4,12 ?

If 144/x is an integer and 108/x is an integer, which of the following must be true?
I. 9/x is an integer
II. 12/x is an integer
III. 36/x is an integer

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

The question asks which of the following MUST be true, not COULD be true. The largest possible value of x is 36 (the greatest common factor of 144 and 108), and if x=36 then ONLY III is true.

Answer: B.
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If \(\frac{144}{x}\) is an integer, and \(\frac{108}{x}\) is an integer, which of the following must be true?


I. \(\frac{9}{x}\) is an integer
II. \(\frac{12}{x}\) is an integer
III. \(\frac{36}{x}\) is an integer


(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III

Source: Gmat Hacks 1800 set.

This is a repost, but the previous postings are still confusing. So the way I read this is "If \(\frac{144}{2}\) is an integer and \(\frac{108}{2}\) is an integer .." but that reasoning seems to be wrong, can someone explain why?

Edit - I get it now. It's the "MUST BE TRUE" part that I forgot to factor into. Anyway, if you get that part, this is pretty easy question.
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can admin please help on this one.........
the way i see is x can be 1 or 2 or 3

how do we decide whether 9/x , 12/x or 36/x is an integer if i chose x as 1 still {144}/{x} is an integer, and {108}/{x} is also an integer...same happens when i chose 2 but x as 1 and x as 2 given me differnt answer choices

I think your doubt is addressed here: if-144-x-is-an-integer-and-108-x-is-an-integer-which-of-the-128415.html#p1058688

x could be 1, 2, 3, 4, 6, 9, 12, 18 or 36 (these are common factors of 144 and 108). The question asks which of the options MUST be an integer. Now, only 36/x is an integer for all possible values of x.

Does this make sense?

Similar questions to practice:
if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html
if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html
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if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html
if-x-and-y-are-positive-integers-and-180x-y-100413.html
n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html
if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html
if-n-and-y-are-positive-integers-and-450y-n-92562.html
if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html

Hope this helps.
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So the best way to start is to break each one into their prime factors:

144=2*2*2*2*3*3

108=2*2*3*3*3

We know that x has to be a combination of the primes such that it is a prime or combination of the primes that it both of those numbers can be evenly divided by it.

Now let's take a look at the options. These are things that MUST be true, so if we can find a scenario where they are not then we know that we can eliminate it.

I. 9/x doesn't have to be an integer because x could be 2 (This eliminates A, C and E)
II. 12/x doesn't have to be an integer because x could be 9 (This eliminates D)

Now we know that B is the only option left we can double check it

III. 36 (2*2*3*3) does have to be an integer because there is no singular or combination of primes that divides evenly into 144 and 108 and not 36.

Thus the answer is B

So we need to find the largest number to X that is divisible by both 144 and 108.

144=3^2 x 4^2
108=3^3 x 4

So looking at this the only numbers that can be divisible is 3^2 x 4 = 36

Looking at the options.

1. 9/36 is not divisible
2. 12/36 is not divisible
3. 36/36 is divisible

Thus the only option that works is 3.

I attacked this question in a different way:
prime factorization of 144 = 2*2*2*2*3*3 = so we have four 4's and two of 3's
prime factorization of 108 = 2*2*3*3*3 = we have two of 2's and three of 3's.

I 9/x is an integer. well, if x is 3*3 = then yes, 9/x is an integer.
but if x is 2*2*3*3 = then 9/x is not divisible. since our question asks for must be true -> we know for sure that I is not true.
Eliminate (A) I only, (C) I and II only, and (E) I, II, and III

II 12/x is an integer
well, if x is 2*2*3 = or 2*2 or 2*3 = then yes, 12/x is an integer, but x can be 3*3*2.
since it is a must be true, we can eliminate E, and thus B is the answer.

B works fine ...

X can be 3, 4, 9, 12 or 36...
hence for any of the above values of x, only 36/x will be an integer ...
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We are given that 144/x is an integer and 108/x is an integer. Let’s prime factorize 144 and 108.

144 = 12 x 12 = 2^2 x 3^1 x 2^2 x 3^1 = 2^4 x 3^2

108 = 4 x 27 = 2^2 x 3^3

Thus, the largest possible value x could be is 2^2 x 3^2 = 4 x 9 = 36, which is the GCF of 144 and 108. Furthermore, x could be any of the factors of 36. Thus, of the Roman numerals, only 36/x must be an integer.

Answer: B
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ScottTargetTestPrep

Could,´t be x=1?

I just have that question-based in your above explanation.

Kind regards!

Yes, x could be 1; however, we care about what MUST be true. That is why we immediately started with the largest possible value of x, which is 36. By doing so, we immediately see that 12/x does not have to be an integer, nor does 9/x, right?
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144/x = Integer
i.e. x is a factor of 144
i.e. x is a factor of \(2^4*3^2\)

108/x = Integer
i.e. x is a factor of 108
i.e. x is a factor of \(2^2*3^3\)

i.e. \(x_{max}\) = HCF of 108 and 144 = \(2^2*3^2 = 36\)

I. 9/x is an integer. NOT NECESSARILY for x = 36
II. 12/x is an integer. NOT NECESSARILY for x = 36
III. 36/x is an integer. ALWAYS TRUE EVEN FOR x=36 (highest value)

Answer: Option B

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