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For how many positive values of n are both \(n\over{2}\) and 3n four-digit base 9 integers?

Best Answer 

 #1
avatar+9664 
+1

\(\frac n2\) is an integer implies n is even. The smallest four-digit base 9 integer is \((1000)_9\), which is \(9^3 = 729\), but this is not even, so \(\frac n2\) is at least 729 + 1 = 730. This gives the lower bound \(n \geq 1460\).

While n/2 is a four-digit base 9 integer, 3n cannot exceed the maximum value of four-digit base 9 integers. that value is \((8888)_9\), which is \(8(9^3 + 9^2 + 9 + 1) = 6560\), but it is not divisible by 3. That means \(3n \leq 6558\), i.e., \(n \leq 2186\) is the upper bound.

 

To ensure that n/2 is an integer, the answer is the number of even numbers n in the range \(1460 \leq n \leq 2186\), i.e., \(\dfrac{2186 - 1460}2 + 1 = \boxed{364}\).

 Apr 8, 2024
 #1
avatar+9664 
+1
Best Answer

\(\frac n2\) is an integer implies n is even. The smallest four-digit base 9 integer is \((1000)_9\), which is \(9^3 = 729\), but this is not even, so \(\frac n2\) is at least 729 + 1 = 730. This gives the lower bound \(n \geq 1460\).

While n/2 is a four-digit base 9 integer, 3n cannot exceed the maximum value of four-digit base 9 integers. that value is \((8888)_9\), which is \(8(9^3 + 9^2 + 9 + 1) = 6560\), but it is not divisible by 3. That means \(3n \leq 6558\), i.e., \(n \leq 2186\) is the upper bound.

 

To ensure that n/2 is an integer, the answer is the number of even numbers n in the range \(1460 \leq n \leq 2186\), i.e., \(\dfrac{2186 - 1460}2 + 1 = \boxed{364}\).

MaxWong Apr 8, 2024
 #2
avatar+9664 
+1

Sorry, there is a mistake in my argument: \(\frac n2\) can in fact be 729, so the lower bound is \(n \geq 1458\). The answer is then 365 instead of 364.

MaxWong  Apr 8, 2024

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