2007 AIME I Problems/Problem 5


The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer.

For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature?


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 $0$.


Proof that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$ iff $k=\left[ \frac{9}{5}x\right]$ 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 celcius 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

See also

2007 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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