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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 Gmat_mission » Sat Aug 22, 2020 4:06 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

Answer: B

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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 priyasinglaa » Sat Aug 22, 2020 10:58 pm
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.
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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 Aman Faruki » Mon Aug 24, 2020 7:01 am
Thanks Priya!
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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 Rinksss » Mon Aug 24, 2020 7:26 am
Thanks
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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 Reenadddd » Mon Aug 24, 2020 7:29 am
Thanks
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