Difference between revisions of "2007 AIME I Problems/Problem 5"

Revision as of 18:29, 11 July 2011

Problem

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

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 $535 + 4 = 539$ 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 $x$ . 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 $\left[968*\frac{5}{9}\right]+1$ , adding in the extra solution corresponding to $0$ .

Solution 3

Let $c$ be a degree Celcius, and $f=\frac 95c+32$ rounded to the nearest integer. $|f-((\frac 95)c+32)|\leq 1/2$ $|(\frac 59)(f-32)-c|\leq \frac 5{18}$ so it must round to $c$ because $\frac 5{18}<\frac 12$ . 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.

@@ Line 9: / Line 9: @@
 Examine <math>F - 32</math> modulo 9.
-*If <math>\displaystyle F - 32 \equiv 0 \pmod{9}</math>, then we can define <math>9x = F - 32</math>. This shows that <math>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</math>. This case works.
+*If <math>F - 32 \equiv 0 \pmod{9}</math>, then we can define <math>9x = F - 32</math>. This shows that <math>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</math>. This case works.
-* If <math>\displaystyle F - 32 \equiv 1 \pmod{9}</math>, then we can define <math>9x + 1 = F - 32</math>. This shows that <math>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</math><math>F = \left[9x + \frac{9}{5}+ 32 \right] \Longrightarrow F = 9x + 34</math>. So this case doesn't work.
+* If <math>F - 32 \equiv 1 \pmod{9}</math>, then we can define <math>9x + 1 = F - 32</math>. This shows that <math>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</math><math>F = \left[9x + \frac{9}{5}+ 32 \right] \Longrightarrow F = 9x + 34</math>. So this case doesn't work.
-Generalizing this, we define that <math>9x + k = F - 32</math>. Thus, <math>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{9}{5}k \right] \right] + 9x + 32</math>. We need to find all values <math>\displaystyle 0 \le k \le 8</math> that <math>\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k</math>. Testing every value of <math>k</math> shows that <math>k = 0, 2, 4, 5, 7</math>, so <math>5</math> of every <math>9</math> values of <math>k</math> work.
+Generalizing this, we define that <math>9x + k = F - 32</math>. Thus, <math>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</math>. We need to find all values <math>0 \le k \le 8</math> that <math>\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k</math>. Testing every value of <math>k</math> shows that <math>k = 0, 2, 4, 5, 7</math>, so <math>5</math> of every <math>9</math> values of <math>k</math> work.
 There are <math>\lfloor \frac{1000 - 32}{9} \rfloor = 107</math> cycles of <math>9</math>, giving <math>5 \cdot 107 = 535</math> numbers that work. Of the remaining <math>6</math> numbers from <math>995</math> onwards, <math>995,\ 997,\ 999,\ 1000</math> work, giving us <math>535 + 4 = 539</math> as the solution.

2007 AIME I (Problems • Answer Key • Resources)
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