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How many positive integers \(n\) have the property that both \(3n\) and \(\dfrac{n}{3}\) are \(4\)-digit integers?

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How many positive integers \(n\) have the property that both \(3n\) and \(\dfrac{n}{3}\) are \(4\)-digit integers?

by M7MBA » Wed May 20, 2020 7:37 am

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How many positive integers \(n\) have the property that both \(3n\) and \(\dfrac{n}{3}\) are \(4\)-digit integers?

A. \(111\)
B. \(112\)
C. \(333 \)
D. \(334\)
E. \(1,134\)

[spoiler]OA=B[/spoiler]

Source: Official Guide
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Re: How many positive integers \(n\) have the property that both \(3n\) and \(\dfrac{n}{3}\) are \(4\)-digit integers?

by Scott@TargetTestPrep » Mon May 25, 2020 10:51 am
M7MBA wrote: ↑
Wed May 20, 2020 7:37 am
 How many positive integers \(n\) have the property that both \(3n\) and \(\dfrac{n}{3}\) are \(4\)-digit integers?

A. \(111\)
B. \(112\)
C. \(333 \)
D. \(334\)
E. \(1,134\)

[spoiler]OA=B[/spoiler]

Source: Official Guide
Solution:

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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