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

#1
+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
+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., $$\dfrac{2186 - 1460}2 + 1 = \boxed{364}$$.

MaxWong Apr 8, 2024
#2
+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