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Senior Manager  G
Joined: 04 Sep 2017
Posts: 291
How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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25 00:00

Difficulty:   95% (hard)

Question Stats: 33% (01:59) correct 67% (02:21) wrong based on 211 sessions

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How many positive integers n have the property that both 3n and n/3 are 4-digit integers?

A. 111
B. 112
C. 333
D. 334
E. 1,134

PS04851.01
Intern  G
Joined: 27 Mar 2019
Posts: 34
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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3
2
Minimum 4 digit number is 1000
Maximum 4 digit number is 9999
max=3n=9999
nmax=3333
min=n3=1000
nmin=3000
Keep in mind that n must be divisible by 3. IMO: the answer would be:
3333−30003+1=112
B

Posted from my mobile device
##### General Discussion
GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5324
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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3
1
4 digit nos
for starts min value 3n at n= 334; 1002 and max at n= 3333 i.e 9999
now the limit given is positive integers n have the property that both 3n and n/3 are 4-digit integers
so n/3 ; 4 digit starts at n=3000 and ends at n=9999
the common range for the given condition is n= 3333-3000; 333 integers i.e values would be 333/3+1 ; 112
IMO b

gmatt1476 wrote:
How many positive integers n have the property that both 3n and n/3 are 4-digit integers?

A. 111
B. 112
C. 333
D. 334
E. 1,134

PS04851.01
Intern  B
Joined: 10 Aug 2019
Posts: 20
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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1
Archit3110 wrote:
4 digit nos
for starts min value 3n at n= 334; 1002 and max at n= 3333 i.e 9999
now the limit given is positive integers n have the property that both 3n and n/3 are 4-digit integers
so n/3 ; 4 digit starts at n=3000 and ends at n=9999
the common range for the given condition is n= 3333-3000; 333 integers i.e values would be 333/3+1 ; 112
IMO b

gmatt1476 wrote:
How many positive integers n have the property that both 3n and n/3 are 4-digit integers?

A. 111
B. 112
C. 333
D. 334
E. 1,134

PS04851.01

Why do we add the 1 in denominator for final answer i.e 333/3+1 ??
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3142
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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3
1

Solution

To find

• The number of integers such that both 3n and n/3 are 4-digit integers.

Approach and Working out

• 1000 < =3n <= 9999
o 1000/3 <= n < = 9999/3
o 333.33 < = n < 3333-----(1)
• 1000 < =n/3 <= 9999
o 3000 < = n < 9999*3---------(2)

• Combining both 1 and 2, we get
o 3000 <= n < 3333
o As n/3 is an integer, n has to be a multiple of 3.
o (3333-3000)/3 +1 = 112
Hence, n can have 112 values.

Thus, option B is the correct answer.

_________________
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3142
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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Vinayak013 wrote:
Archit3110 wrote:
4 digit nos
for starts min value 3n at n= 334; 1002 and max at n= 3333 i.e 9999
now the limit given is positive integers n have the property that both 3n and n/3 are 4-digit integers
so n/3 ; 4 digit starts at n=3000 and ends at n=9999
the common range for the given condition is n= 3333-3000; 333 integers i.e values would be 333/3+1 ; 112
IMO b

gmatt1476 wrote:
How many positive integers n have the property that both 3n and n/3 are 4-digit integers?

A. 111
B. 112
C. 333
D. 334
E. 1,134

PS04851.01

Why do we add the 1 in denominator for final answer i.e 333/3+1 ??

We add +1 because when we subtract two numbers, let's say 3 and 1. The result is the gap between them. 3-1 = 2 (Gap between 3 and 1)

So, if we want to include all the three numbers, we need to add 1 to it.

I hope this helps you.
Regards,
Ashutosh
e-GMAT Quant Expert
_________________
Intern  B
Joined: 10 Aug 2019
Posts: 20
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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Still not clear can you Please elaborate?
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3142
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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1
Hey @Vinayaka,

Okay, let's answer a few of my question and i am sure you will understand what I am trying to say.

How many numbers are there from 1 to 10, both inclusive?
Is the answer equal to 10-1?
Is the answer equal to (10-1) +1?

You will find that answer is equal to (10-1)+1 and this is because the difference between two numbers gives the gap between two numbers and, total consecutive number that forms the gaps is always 1 greater than the gap.

So, between 1 and 10, there are 9 gaps and it formed by 10 consecutive numbers.

I hope this helps you.
Regards,
_________________
Manager  P
Joined: 01 Feb 2017
Posts: 244
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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1
n= x/3 = y*3 where x and y are 4 digit integers.

Therefore, x = 9*y

Min y = 1000, Corresponding Min x= 9000
Max x= 9999, Corresponding Max y= 1111

Possible values of n= (1111-1000) + 1 = 112 OR (9999-9000)/9 + 1 = 112

Ans B
Intern  B
Joined: 10 Aug 2019
Posts: 20
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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EgmatQuantExpert wrote:
Hey @Vinayaka,

Okay, let's answer a few of my question and i am sure you will understand what I am trying to say.

How many numbers are there from 1 to 10, both inclusive?
Is the answer equal to 10-1?
Is the answer equal to (10-1) +1?

You will find that answer is equal to (10-1)+1 and this is because the difference between two numbers gives the gap between two numbers and, total consecutive number that forms the gaps is always 1 greater than the gap.

So, between 1 and 10, there are 9 gaps and it formed by 10 consecutive numbers.

I hope this helps you.
Regards,

Got it , Thanks a ton!
Manager  B
Joined: 02 Nov 2018
Posts: 53
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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EgmatQuantExpert wrote:

Solution

To find

• The number of integers such that both 3n and n/3 are 4-digit integers.

Approach and Working out

• 1000 < =3n <= 9999
o 1000/3 <= n < = 9999/3
o 333.33 < = n < 3333-----(1)
• 1000 < =n/3 <= 9999
o 3000 < = n < 9999*3---------(2)

• Combining both 1 and 2, we get
o 3000 <= n < 3333
o As n/3 is an integer, n has to be a multiple of 3.
o (3333-3000)/3 +1 = 112
Hence, n can have 112 values.

Thus, option B is the correct answer.

If 3000 <= n < 3333 then why isn't the answer 334? There are 334 integers in this range..
VP  D
Joined: 14 Feb 2017
Posts: 1291
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GPA: 3
WE: Management Consulting (Consulting)
Re: How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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jamalabdullah100 wrote:
EgmatQuantExpert wrote:

Solution

To find

• The number of integers such that both 3n and n/3 are 4-digit integers.

Approach and Working out

• 1000 < =3n <= 9999
o 1000/3 <= n < = 9999/3
o 333.33 < = n < 3333-----(1)
• 1000 < =n/3 <= 9999
o 3000 < = n < 9999*3---------(2)

• Combining both 1 and 2, we get
o 3000 <= n < 3333
o As n/3 is an integer, n has to be a multiple of 3.
o (3333-3000)/3 +1 = 112
Hence, n can have 112 values.

Thus, option B is the correct answer.

If 3000 <= n < 3333 then why isn't the answer 334? There are 334 integers in this range..

We obtained the more defined range be looking at potential values for 3n and n/3.

Because n/3 must be an integer we are only concerned with integers divisible by 3, so we only count those integers.

We count integers divisible by a specific integer such as 3 by (upper range - lower range)/multiple we try to count + 1
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Manager  S
Joined: 29 Oct 2015
Posts: 233
How many positive integers n have the property that both 3n and n/3 ar  [#permalink]

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EgmatQuantExpert wrote:

Solution

To find

• The number of integers such that both 3n and n/3 are 4-digit integers.

Approach and Working out

• 1000 < =3n <= 9999
o 1000/3 <= n < = 9999/3
o 333.33 < = n < 3333-----(1)
• 1000 < =n/3 <= 9999
o 3000 < = n < 9999*3---------(2)

• Combining both 1 and 2, we get
o 3000 <= n < 3333
o As n/3 is an integer, n has to be a multiple of 3.
o (3333-3000)/3 +1 = 112
Hence, n can have 112 values.

Thus, option B is the correct answer.

Hey egmat
EgmatQuantExpert

You missed the '=' sign here.
I think it should be
333.33 < = n <= 3333-----(1)[/list]

And you missed the '=' sign here also.
IMO it should be
3000 < = n <= 9999*3

Again you missed it here.
It should be ,

Combining both 1 and 2, we get
[list]o 3000 <= n < = 3333

Please check once.  How many positive integers n have the property that both 3n and n/3 ar   [#permalink] 28 Oct 2019, 15:15
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