How many positive integers n have the property that both 3n

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Thu Dec 26, 2019 10:50 pm

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

OA B

Source: Official Guide

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Source: Beat The GMAT — Problem Solving | Thu Dec 26, 2019 10:50 pm

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8154; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sat Dec 28, 2019 8:06 pm

BTGmoderatorDC 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

OA B

Source: Official Guide

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

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deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Thu Jan 02, 2020 5:36 pm

Question: Find the numbers of integers such that both 5n and n/3 are 4-digit integers.
Minimum 4-digit integer = 1000
Maximum 4-digit integer = 9999
$$Therefore,\ \frac{1000}{3}\le\frac{3n}{3}\le\frac{9999}{3}$$
$$333.33\le n\le3333\ \ \ \ ---eqn\ \left(1\right)$$
$$1000\le\frac{n}{3}\le9999$$
$$1000\cdot3\le\frac{n}{3}\cdot3\le9999\cdot3$$
$$3000\le n\le29,997\ ---eqn\left(2\right)$$
Combining both equation together,
Since 333.33 and 29,997 are not 4-digit integers, we will have $$3000\le n\le3333$$
For 'n/3' to be an integer, it must be divisible by 3. So, we use
$$=\frac{upper\ range-lower\ range}{3}+1$$
$$=\frac{3333-3000}{3}+1$$
$$=\frac{333}{3}+1$$

Therefore, the correct answer is option B.

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