How many positive integers n have the property that both 3n and n/3 ar

Question

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

EgmatQuantExpert · Accepted Answer

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.

Correct Answer: Option B

cantaffordname · Answer

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

Archit3110 · Answer

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

Vinayak013 · Answer

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

EgmatQuantExpert · Answer

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

Vinayak013 · Answer

Still not clear can you Please elaborate?

EgmatQuantExpert · Answer

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 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,

Shobhit7 · Answer

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

jamalabdullah100 · Answer

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

dcummins · Answer

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

ScottTargetTestPrep · Answer

Since n/3 is an integer, n must be a multiple of 3. The largest multiple of 3 that contains the mentioned properties is 3333, and the smallest is 3000. The number of multiples of 3 from 3000 to 3333 is:

(3333 - 3000)/3 + 1 = 333/3 + 1 = 112.

Answer: B

KarishmaB · Answer

This is how I thought about it: 3n and n/3 both need to be 4 digit integers so n can be neither 3 digit (n/3 won't be 4 digit) nor 5 digit (3n won't be 4 digit).

So look at 1000, the first 4 digit number. 3n will be 3000 but n/3 will be a 3 digit number.
So then 3000 is the smallest 4 digit number such that n/3 will be 4 digit. Ofcourse, 3n is 9000 which is also 4 digit. The largest possible value of 3n will be 9999 so the largest possible value of n is 3333.

Since n/3 needs to be an integer, n must be a multiple of 3. 
3000 = 3 * 1000
3333 = 3 * 1111

From 000 to 111, these are 112 numbers.

Answer (B)

johnnymbikes · Answer

Pretty good solutions from the experts already but here's a more detailed approach. This is essentially a complex arithmetic sequence question.

Question is asking for 4 digit numbers: _ _ _ _ 
so the smallest # would be 1,000 and the largest would be 9,999

Info #1: 
1,000 <_ 3n <_ 9,999
so: 333.33 <_ n <_ 3,333 
so from here we know that we are looking at only 4 digit numbers so would bump up the range to min: 1,000 <_ n <_ 3,333 (ignore the numbers from 333.33 to 999)

Info #2: 
1,000 <_ 1/3n <_ 9,999
3,000 <_ n < 9,999*3 
From here we can infer that 9,999*3 = 29,997 which is 5 digits, and we already knew that that 9,999 was the top of the range so can ignore the 29,997.

Combine the restrictions from Info 1 (max range is 3,333) and Info 2 (min range is 3,000) and we can get that: 
3,000<_ n <_ 3,333

Next, we are looking at the number of multiples of 3 between 3,000 and 3,333 inclusive --> this is an arithmetic sequence.

Apply the formula: An = Ai + (n-1)d 
3,333 = 3000 + (n-1)3 
333 = (n-1)3
111 = n-1
.'. n = 112

23ss23 · Answer

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

Sol: Let's start with the first condition - 3n has to be a 4-digit integer. The largest 4 digit integer is 9999, hence the highest value in our solution range can be 9999/3 = 3333.

Now the second condition - n/3 has to be a 4-digit integer. The smallest 4 digit integer is 1000, hence the smallest value in our solution ranger can be 1000*3 = 3000.

The range of value we currently have is 3333-3000 = 333 + 1 (+1 since 3333 and 3000, both are included). But these values should be divisible by 3. Out of the given range of 334 numbers, we know that one in every three numbers is divisible by 3. If we start with 3000, which is divisible by 3, we end up having 112 values which satisfy the condition in the range the values (3000 - 3333).

pappal · Answer

the first no. is 3000: 3000/3=1000 and 3000*3=9000
the last no. is 3333: 3333/3=1111 and 3333*3=9999
and all the no.'s between 3000 and 3333 should be in AP with common difference=3
so we can have 3000+(n-1)3=3333
=>3n-3=333
=>3n=336
n=112 so B

Kinshook · Answer

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

n/3 >= 1000; n>=3000
3n <= 9999; n<=3333

Common difference = 3

Number of positive integers n = (3333-3000)/3 + 1 = 112

IMO B

How many positive integers n have the property that both 3n and n/3 ar

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GMAT Problem Solving (PS) Questions

How many positive integers n have the property that both 3n and n/3 ar

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