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If n is a positive integer and n^2 is divisible by 72, then

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If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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17 Mar 2014, 00:16
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

Problem Solving
Question: 169
Category: Arithmetic Properties of numbers
Page: 84
Difficulty: 600

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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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17 Mar 2014, 00:16
9
22
SOLUTION

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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17 Mar 2014, 04:51
6
1
$$\frac{n*n}{72}$$ = $$\frac{n*n}{2*2*2*3*3}$$
This means n should be at least 2*2*3. Otherwise $$n^2$$ won't be divisible by 72

What is the largest integer that can divide n?
Let's see, largest integer that can divide 2? 2
Largest integer that can divide 3? 3
Largest integer that can divide n is n itself

Time Taken 2:31
Difficulty level 650
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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18 Mar 2014, 02:10
1

We require to find the least value of n

Say n=6, however 36 is not divisible by 72; Ignore A

Say n=12, 144 is divisible by 72

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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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Updated on: 23 Mar 2014, 20:01
2
1
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

Given that, n is a positive integer and n^2 is divisible by 72 means that n^2 is a perfect square that is lowest multiple of 72.

Or, the lowest multiple of 72 that is a perfect square is 144.

So, n^2 = 144q, q is some positive integer => 12 is a factor of n.

Originally posted by arunspanda on 18 Mar 2014, 08:00.
Last edited by arunspanda on 23 Mar 2014, 20:01, edited 2 times in total.
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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23 Mar 2014, 07:57
SOLUTION

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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28 Jul 2014, 10:02
Bunuel wrote:
SOLUTION

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

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Hi,

How can the part highlighted in bold be concluded fro the question stem "The largest positive integer that must divide n, means for the least value of n"??

In absence of this even 48 looks like a valid answer, as 48^2 is divisible by 72 and 48 is divisible by itself.

Regards,
Sagar
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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28 Jul 2014, 10:12
3
itsworththepain wrote:
Bunuel wrote:
SOLUTION

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

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Hi,

How can the part highlighted in bold be concluded fro the question stem "The largest positive integer that must divide n, means for the least value of n"??

In absence of this even 48 looks like a valid answer, as 48^2 is divisible by 72 and 48 is divisible by itself.

Regards,
Sagar

We need the largest positive integer that MUST divide n. So, we should find the least value of n for which n^2 is divisible by 72, and if that n is divisible by some number then so will be every other n's (MUST condition will be satisfied). The least positive n for which n^2 is divisible by 72 is 12 (12^2 = 144, which is divisible by 72). So, even if n = 12 it is divisible by 12 but not divisible by 24, 36 or 48.

Follow the links in my previous post for similar questions to understand the concept better.
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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14 Mar 2015, 08:56
1
Bunuel wrote:
itsworththepain wrote:
Bunuel wrote:
SOLUTION

We need the largest positive integer that MUST divide n. So, we should find the least value of n for which n^2 is divisible by 72, and if that n is divisible by some number then so will be every other n's (MUST condition will be satisfied). The least positive n for which n^2 is divisible by 72 is 12 (12^2 = 144, which is divisible by 72). So, even if n = 12 it is divisible by 12 but not divisible by 24, 36 or 48.

Follow the links in my previous post for similar questions to understand the concept better.

I don't actually understand this either, why is it implied that we need the least value for n? I mean, as the previous poster said. 48^2 too, is divisible by 72 and 48/48 = 1.

Why is it (I mean, how do I interpret from the question that we need the least value for n)
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If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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Updated on: 03 Nov 2015, 01:01
1
Let's make it look easy

72= 2*2*2*3*3 we need one more 2 in order to make a perfect square we need an even number of primes --> we need one (2)
--> 72*2 =144, Take a square root of it an you'll get 12 (B)

Originally posted by BrainLab on 21 Jun 2015, 05:16.
Last edited by BrainLab on 03 Nov 2015, 01:01, edited 1 time in total.
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If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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21 Jun 2015, 07:35
1
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

CONCEPT: Perfect Squares always have even power of Prime no. in their Prime factorised form.

$$72 = 8*9 = 2^3*3^2$$

Since power of prime numbers must be prime so $$72$$must be multiplied by minimum $$2^1$$ to make the result a perfect Square

In that Case $$n^2 = 2^4 * 3^2$$

i.e. $$n = 2^2 * 3 = 12$$

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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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22 Nov 2015, 20:34
I did it with exponents as other people have done.
72 = 9*8 = 3^2*2^3

n^2=72x
n = 3*2^2 (x) and X could be anything positive but 12 MUST divide n
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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23 Nov 2015, 03:24
I would like to emphasize one point here (which I think is not very explicit in previous answers). The following step is straightforward,
n^2 = 72 M = 2*6*6*M, where M is an integer.
For perfect square, even prime factors are needed. That is, n^2= 2*6*6*2*m (for some integer m different from M)
Since, n>0, n= (2*6)*k , where k is some other integer. That implies, n=12k where k is an integer. Meaning, n is not just a single number but any multiple of 12. This is where the question phrase, "then the largest positive integer that must divide n is.." plays the part. Since k= 1 or 2 or 3 or any other integer, no matter how large the number, "12" would be the number that would always divide "n". 6 would do the same (so would 3 or 4 or 2 hypothetically). But the largest one to always divide "n" would be "12" [among 1, 2, 3, 4, 6 or 12].
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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18 Apr 2017, 07:43
Option B

$$n^2 = 72x = 2^3*3^2*x = 2^4*3^2*w$$ : A perfect square contains Even powers of all the prime factors. [x, w & k are integers]
$$n = 2^2*3*k = 12k$$
$$k = \frac{n}{12}$$ = integer.

12 is the largest integer which must divide n evenly.
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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22 Sep 2017, 11:49
72 = 2^3 * 3^2

Given that n^2 is divisible by 72. So in order to satisfy this condition, n has to have a minimum value of 2^2 * 3^2 squaring which would be divisible by 72.
Hence n = 12 and largest possible value that can divide n is 12.

Kudos if it helps
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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11 May 2018, 11:01
1
Bunuel wrote:
itsworththepain wrote:
Bunuel wrote:
SOLUTION

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

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Hi,

How can the part highlighted in bold be concluded fro the question stem "The largest positive integer that must divide n, means for the least value of n"??

In absence of this even 48 looks like a valid answer, as 48^2 is divisible by 72 and 48 is divisible by itself.

Regards,
Sagar

We need the largest positive integer that MUST divide n. So, we should find the least value of n for which n^2 is divisible by 72, and if that n is divisible by some number then so will be every other n's (MUST condition will be satisfied). The least positive n for which n^2 is divisible by 72 is 12 (12^2 = 144, which is divisible by 72). So, even if n = 12 it is divisible by 12 but not divisible by 24, 36 or 48.

Follow the links in my previous post for similar questions to understand the concept better.

niks18 hello

is there a way to rephrase ths sentence so as i can understand

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question.

it says " largest positive integer " and at the same time mentiones "the least value"
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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11 May 2018, 20:28
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

let n^2=72x
72=6*6*2
to get two matched factor pairs,
let x=2
n^2=6*6*2*2=144
√144=n=12
B
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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14 May 2018, 07:32
1
1
dave13 wrote:
Bunuel wrote:
Bunuel wrote:
SOLUTION

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is

(A) 6
(B) 12
(C) 24
(D) 36
(E) 48

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question. The lowest square of an integer, which is multiple of $$72$$ is $$144$$ --> $$n^2=144=12^2=72*2$$ --> $$n_{min}=12$$. Largest factor of $$12$$ is $$12$$.

OR:

Given: $$72k=n^2$$, where $$k$$ is an integer $$\geq1$$ (as $$n$$ is positive).

$$72k=n^2$$ --> $$n=6\sqrt{2k}$$, as $$n$$ is an integer $$\sqrt{2k}$$, also must be an integer. The lowest value of $$k$$, for which $$\sqrt{2k}$$ is an integer is when $$k=2$$ --> $$\sqrt{2k}=\sqrt{4}=2$$ --> $$n=6\sqrt{2k}=6*2=12$$

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We need the largest positive integer that MUST divide n. So, we should find the least value of n for which n^2 is divisible by 72, and if that n is divisible by some number then so will be every other n's (MUST condition will be satisfied). The least positive n for which n^2 is divisible by 72 is 12 (12^2 = 144, which is divisible by 72). So, even if n = 12 it is divisible by 12 but not divisible by 24, 36 or 48.

Follow the links in my previous post for similar questions to understand the concept better.

niks18 hello

is there a way to rephrase ths sentence so as i can understand

The largest positive integer that must divide $$n$$, means for the least value of $$n$$ which satisfies the given statement in the question.

it says " largest positive integer " and at the same time mentiones "the least value"

Hi dave13

you need to note few things here -

$$n^2$$ will always be divisible by 72, so whatever the value of n is, its square has to be a multiple of 72

therefore $$n^2=72k$$, for some constant k

$$=>n=6\sqrt{2k}$$, now as $$n$$ is integer, hence $$2k$$ has to be a perfect square, so let's assume $$k=2x^2$$, for some integer $$x$$

$$=>n=6*2x=12x$$

Now look at the options given: 6, 12, 24, 36, 48. 6 is not possible because least value of $$n$$ is 12 and 12>6. so 12 is the definite contender to be the largest divisor on $$n$$

why 24, 36 & 48 are not possible here? we have got $$n=12x$$, so essentially value of $$n$$ will depend on integer $$x$$

if x=1, $$n$$ will not be divisible by 24, 36 & 48. if x=2, then $$n$$ will not be divisible by 36 & 48. if x=3, $$n$$ will not be divisible by 24 & 48 and so on.

Hence $$n$$ has to be 12 for $$n^2$$ to be ALWAYS divisible by 72. This is a MUST BE TRUE question. Hence we need a definite value here.

So from the above explanation, we have derived, that "The largest integer that must divide $$n$$, for $$n^2$$ to be always divisible by 72", we actually need the LEAST value of $$n$$
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Re: If n is a positive integer and n^2 is divisible by 72, then  [#permalink]

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18 May 2019, 01:24
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Re: If n is a positive integer and n^2 is divisible by 72, then   [#permalink] 18 May 2019, 01:24
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