AIME 2007 I · 第 5 题

Solution

Solution 1

Examine $F - 32$ modulo 9.

• If $F - 32 \equiv 0 \pmod{9}$, then we can define $9x = F - 32$. This shows that $F = \left[\frac{9}{5}\left[\frac{5}{9}(F-32)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x) + 32\right] \Longrightarrow F = 9x + 32$. This case works.
• If $F - 32 \equiv 1 \pmod{9}$, then we can define $9x + 1 = F - 32$. This shows that $F = \left[\frac{9}{5}\left[\frac{5}{9}(F-32)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + 1) + 32\right] \Longrightarrow$$F = \left[9x + \frac{9}{5}+ 32 \right] \Longrightarrow F = 9x + 34$. So this case doesn't work.

Generalizing this, we define that $9x + k = F - 32$. Thus, $F = \left[\frac{9}{5}\left[\frac{5}{9}(9x + k)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + \left[\frac{5}{9}k\right]) + 32\right] \Longrightarrow F = \left[\frac{9}{5} \left[\frac{5}{9}k \right] \right] + 9x + 32$. We need to find all values $0 \le k \le 8$ that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$. Testing every value of $k$ shows that $k = 0, 2, 4, 5, 7$, so $5$ of every $9$ values of $k$ work.

There are $\lfloor \frac{1000 - 32}{9} \rfloor = 107$ cycles of $9$, giving $5 \cdot 107 = 535$ numbers that work. Of the remaining $6$ numbers from $995$ onwards, $995,\ 997,\ 999,\ 1000$ work, giving us $535 + 4 = \boxed{539}$ as the solution.

Solution 2

Notice that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$ holds if $k=\left[ \frac{9}{5}x\right]$ for some integer $x$. Thus, after translating from $F\to F-32$ we want count how many values of $x$ there are such that $k=\left[ \frac{9}{5}x\right]$ is an integer from $0$ to $968$. This value is computed as $\left[968*\frac{5}{9}\right]+1 = \boxed{539}$, adding in the extra solution corresponding to $0$.

Note

Proof that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$ if $k=\left[ \frac{9}{5}x\right]$ for some integer $x$:

First assume that $k$ cannot be written in the form $k=\left[ \frac{9}{5}x\right]$ for any integer $x$. Let $z = \left[ \frac{5}{9}k\right]$. Our equation simplifies to $k = \left[ \frac{9}{5}z\right]$. However, this equation is not possible, as we defined $k$ such that it could not be written in this form. Therefore, if $k \neq \left[ \frac{9}{5}x\right]$, then $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] \neq k$.

Now we will prove that if $k = \left[ \frac{9}{5}x\right]$, $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$. We realize that because of the 5 in the denominator of $\left[ \frac{9}{5}x \right]$, $\left[ \frac{9}{5}x \right]$ will be at most $\frac{2}{5}$ away from $\frac{9}{5}x$. Let $z = \left[ \frac{9}{5}x \right]- \frac{9}{5}x$, meaning that $-\frac{2}{5} \leq z \leq \frac{2}{5}$. Now we substitute this into our equation:

$$\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9} \left[ \frac{9}{5}x\right] \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9} (\frac{9}{5}x + z) \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9} (\frac{9}{5}x + z) \right] \right] = \left[ \frac{9}{5} \left[ x+ \frac{5}{9}z \right] \right]$$ .

Now we use the fact that $-\frac{2}{5} \leq z \leq \frac{2}{5}$

$$\left[ \frac{9}{5} \left[ x - \frac{5}{9}(\frac{2}{5}) \right] \right] \leq \left[ \frac{9}{5} \left[ x + \frac{5}{9}(z) \right] \right] \leq \left[ \frac{9}{5} \left[ x + \frac{5}{9}(\frac{2}{5}) \right] \right]$$ $$\left[ \frac{9}{5} x \right] \leq \left[ \frac{9}{5} \left[ x + \frac{5}{9}(z) \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9}k \right] \right] \leq \left[ \frac{9}{5}x \right]$$ Hence $\left[ \frac{9}{5} \left[ \frac{5}{9}k \right] \right] = \left[ \frac{9}{5}x \right] = k$ and we are done.

- mako17

Solution 3

Let $c$ be a degree Celsius, and $f=\frac 95c+32$ rounded to the nearest integer. Since $f$ was rounded to the nearest integer we have $|f-((\frac 95)c+32)|\leq 1/2$, which is equivalent to $|(\frac 59)(f-32)-c|\leq \frac 5{18}$ if we multiply by $5/9$. Therefore, it must round to $c$ because $\frac 5{18}<\frac 12$ so $c$ is the closest integer. Therefore there is one solution per degree Celsius in the range from $0$ to $(\frac 59)(1000-32) + 1=(\frac 59)(968) + 1=538.8$, meaning there are $539$ solutions.

Solution 4

Start listing out values for $F$ and their corresponding values of $C$. You will soon find that every 9 values starting from $F$ = 32, there is a pattern:

$F=32$: Works

$F=33$: Doesn't work

$F=34$: work

$F=35$: Doesn’t work

$F=36$: Works

$F=37$: Works

$F=38$: Doesn’t work

$F=39$: Works

$F=40$: Doesn’t work

$F=41$: Works

There are $969$ numbers between $32$ and $1000$, inclusive. This is $107$ sets of $9$, plus $6$ extra numbers at the end. In each set of $9$, there are $5$ “Works,” so we have $107\cdot5 = 535$ values of $F$ that work.

Now we must add the $6$ extra numbers. The number of “Works” in the first $6$ terms of the pattern is $4$, so our final answer is $535 + 4 = 539$ solutions that work.

Submitted by warriorcats

Solution 5(similar to solution 3 but faster solution if you have no time)

Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $\boxed{539}$ solutions.

-alanisawesome2018

Video Solution

2007 AIME I #5

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