AIME 2007 II · 第 12 题

Solution

Suppose that $x_0 = a$, and that the common ratio between the terms is $r$.

The first conditions tells us that $\log_3 a + \log_3 ar + \ldots + \log_3 ar^7 = 308$. Using the rules of logarithms, we can simplify that to $\log_3 a^8r^{1 + 2 + \ldots + 7} = 308$. Thus, $a^8r^{28} = 3^{308}$. Since all of the terms of the geometric sequence are integral powers of $3$, we know that both $a$ and $r$ must be powers of 3. Denote $3^x = a$ and $3^y = r$. We find that $8x + 28y = 308$. The possible positive integral pairs of $(x,y)$ are $(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)$.

The second condition tells us that $56 \le \log_3 (a + ar + \ldots ar^7) \le 57$. Using the sum formula for a geometric series and substituting $x$ and $y$, this simplifies to $3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}$. The fractional part $\approx \frac{3^{8y}}{3^y} = 3^{7y}$. Thus, we need $\approx 56 \le x + 7y \le 57$. Checking the pairs above, only $(21,5)$ is close.

Our solution is therefore $\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}$.

Solution 2

All these integral powers of $3$ are all different, thus in base $3$ the sum of these powers would consist of $1$s and $0$s. Thus the largest value $x_7$ must be $3^{56}$ in order to preserve the givens. Then we find by the given that $x_7x_6x_5\dots x_0 = 3^{308}$, and we know that the exponents of $x_i$ are in an arithmetic sequence. Thus $56+(56-r)+(56-2r)+\dots +(56-7r) = 308$, and $r = 5$. Thus $\log_3 (x_{14}) = \boxed{091}$.

Solution 3

Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call $x_0$, $x_1$, $x_2$..., as $3^n$, $3^{n+m}$, and $3^{n+2m}$... respectively. With this format we can rewrite the first given equation as $n + n + m + n+2m + n+3m+...+n+7m = 308$. Simplify to get $2n+7m=77$. (1)

Now, rewrite the second given equation as $3^{56} \leq \left( \sum_{n=0}^{7}x_{n} \right) \leq 3^{57}$. Obviously, $x_7$, aka $3^{n+7m}$ $<3^{57}$ because there are some small fractional change that is left over. This means $n+7m$ is $\leq56$. Thinking about the geometric sequence, it's clear each consecutive value of $x_i$ will be either a power of three times smaller or larger. In other words, the earliest values of $x_i$ will be negligible compared to the last values of $x_i$. Even in the best case scenario, where the common ratio is 3, the values left of $x_7$ are not enough to sum to a value greater than 2 times $x_7$ (amount needed to raise the power of 3 by 1). This confirms that $3^{n+7m} = 3^{56}$. (2)

Use equations 1 and 2 to get $m=5$ and $n=21$. $\log_{3}{x_{14}} = \log_{3}{3^{21+14*5}} = 21+14*5 = \boxed{091}$

-jackshi2006

Solution 4 (dum)

Proceed as in Solution 3 for the first few steps. We have the sequence $3^{a},3^{a+n},3^{a+2n}...$. As stated above, we then get that $a+a+n+...+a+7n=308$, from which we simplify to $2a+7n=77$. From here, we just go brute force using the second statement (that $3^{56}\leq 3^{a}+...+3^{a+7n}\leq 3^{57}$). Rearranging the equation from earlier, we get

$$n=11-\frac{2a}{7}$$ from which it is clear that $a$ is a multiple of $7$. Testing the first few values of $a$, we get: Case 1 ($a=7,n=9$) The sequence is then $3^{7}+...+3^{70}$, which breaks the upper bound. Case 2 ($a=14,n=7$) The sequence is then $3^{14}+...+3^{63}$, which also breaks the upper bound. Case 3 ($a=21,n=5$) This is the first reasonable one, giving $3^{21}+...+3^{56}$. It seems like this would break the upper bound, but from some testing we get:

$$3^{21}+...+3^{56}<3^{57}$$ $$1+...+3^{35}<3^{36}$$ $$1+...+3^{30}<2*3^{35}$$ $$3^{5}+...+3^{30}<2*3^{35}-1$$ $$1+...+3^{25}<2*3^{30}-3^{-5}$$ Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over)

$$1<2*3^{5}$$ which confirms that this satisfies our upper bound. Thus $a=21,n=5$, so $x_14=3^{a+14n}\rightarrow3^{91}$. We then get the requested answer, $\log_3(3^{91})=\boxed{091}$ ~ amcrunner

Solution 5

Let $\log_3(x_n)=a_n$. Our goal is to find $a_{14}$. Since the sequence consists of integral powers of $3$, all $a_n$ are integers.

From the first given condition, we have

$$a_0+a_1+a_2+\cdots+a_7=308.$$ From the second condition, we claim that $a_7=56$. Since all $x_n$ are powers of $3$, the largest possible values of the eight terms are $3^{56},3^{55},3^{54},\dots,3^{49}$. Even if these formed an infinite geometric series with first term $3^{56}$ and ratio $\frac13$, the sum would be $\frac{3^{57}}{2}$, whose base-$3$ logarithm lies strictly between $56$ and $57$. Since

$$\sum_{n=0}^{7} x_n < \sum_{n=0}^{\infty} 3^{56}\left(\frac13\right)^n,$$ it follows that $x_7=3^{56}$, so $a_7=56$.

Let $d$ denote the common difference of the arithmetic sequence $\{a_n\}$. Then the first condition can be rewritten as

$$56+(56-d)+(56-2d)+\cdots+(56-7d)=308.$$ Simplifying,

$$8\cdot 56 - 28d = 308,$$ so

$$-28d=-140 \quad \text{and} \quad d=5.$$ Therefore, $a_{14}=56+7\cdot 5=\boxed{091}.$

~Voidling