Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 04 Sep 2017
Posts: 291

How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
21 Sep 2019, 13:53
Question Stats:
33% (02:00) correct 67% (02:20) wrong based on 198 sessions
HideShow timer Statistics
How many positive integers n have the property that both 3n and n/3 are 4digit integers? A. 111 B. 112 C. 333 D. 334 E. 1,134 PS04851.01
Official Answer and Stats are available only to registered users. Register/ Login.




Intern
Joined: 27 Mar 2019
Posts: 34

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
21 Sep 2019, 14:00
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




GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5271
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]
Show Tags
23 Sep 2019, 04:34
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 4digit integers so n/3 ; 4 digit starts at n=3000 and ends at n=9999 the common range for the given condition is n= 33333000; 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 4digit integers?
A. 111 B. 112 C. 333 D. 334 E. 1,134
PS04851.01



Intern
Joined: 10 Aug 2019
Posts: 19

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
28 Sep 2019, 07:29
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 4digit integers so n/3 ; 4 digit starts at n=3000 and ends at n=9999 the common range for the given condition is n= 33333000; 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 4digit 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 ??



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3138

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
29 Sep 2019, 08:27
Solution To find• The number of integers such that both 3n and n/3 are 4digit 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 (33333000)/3 +1 = 112 Hence, n can have 112 values. Thus, option B is the correct answer. Correct Answer: Option B
_________________



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3138

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
29 Sep 2019, 09:20
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 4digit integers so n/3 ; 4 digit starts at n=3000 and ends at n=9999 the common range for the given condition is n= 33333000; 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 4digit 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. 31 = 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 eGMAT Quant Expert
_________________



Intern
Joined: 10 Aug 2019
Posts: 19

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
29 Sep 2019, 11:06
Still not clear can you Please elaborate?



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3138

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
29 Sep 2019, 14:32
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 10? Is the answer equal to 101? Is the answer equal to (101) +1? You will find that answer is equal to (101)+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
Joined: 01 Feb 2017
Posts: 244

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
30 Sep 2019, 06:38
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= (11111000) + 1 = 112 OR (99999000)/9 + 1 = 112
Ans B



Intern
Joined: 10 Aug 2019
Posts: 19

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
04 Oct 2019, 05:09
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 10? Is the answer equal to 101? Is the answer equal to (101) +1?
You will find that answer is equal to (101)+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
Joined: 02 Nov 2018
Posts: 53

Re: How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
17 Oct 2019, 15:16
EgmatQuantExpert wrote: Solution To find• The number of integers such that both 3n and n/3 are 4digit 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 (33333000)/3 +1 = 112 Hence, n can have 112 values. Thus, option B is the correct answer. Correct Answer: Option BIf 3000 <= n < 3333 then why isn't the answer 334? There are 334 integers in this range..



VP
Joined: 14 Feb 2017
Posts: 1275
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]
Show Tags
18 Oct 2019, 17:45
jamalabdullah100 wrote: EgmatQuantExpert wrote: Solution To find• The number of integers such that both 3n and n/3 are 4digit 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 (33333000)/3 +1 = 112 Hence, n can have 112 values. Thus, option B is the correct answer. Correct Answer: Option BIf 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
_________________
Goal: Q49, V41
+1 Kudos if I have helped you



Manager
Joined: 29 Oct 2015
Posts: 233

How many positive integers n have the property that both 3n and n/3 ar
[#permalink]
Show Tags
28 Oct 2019, 15:15
EgmatQuantExpert wrote: Solution To find• The number of integers such that both 3n and n/3 are 4digit 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 (33333000)/3 +1 = 112 Hence, n can have 112 values. Thus, option B is the correct answer. Correct Answer: Option BHey egmatEgmatQuantExpertYou 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*3Again you missed it here. It should be , Combining both 1 and 2, we get [list]o 3000 <= n < = 3333Please check once.




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






