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

Revision as of 18:44, 15 March 2007

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 $\displaystyle F - 32 \equiv 0 \pmod{9}$ , 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 $\displaystyle F - 32 \equiv 1 \pmod{9}$ , 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, $F = \left[\frac{9}{5}\left[\frac{5}{9}(9x + k)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + \left[\frac{5}{9}k\right]) + 32\right] \Longrightarrow F = \left[\frac{9}{5} \left[\frac{9}{5}k \right] \right] + 9x + 32$ . We need to find all values $\displaystyle 0 \le k \le 8$ 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

Hint. Consider the identity $Round(ax)=Round(xRound(a/Round(ax))$ its something like that...

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 20: / Line 20: @@
 Hint. Consider the identity <math>Round(ax)=Round(xRound(a/Round(ax))</math> its something like that...
 == Solution 3 ==
-A full solution:
+Let <math>c</math> be a degree Celcius, and <math>f=\frac 95c+32</math> rounded to the nearest integer. <math>|f-((\frac 95)c+32)|\leq 1/2</math> <math>|(\frac 59)(f-32)-c|\leq \frac 5{18}</math> so it must round to <math>c</math> because <math>\frac 5{18}<\frac 12</math>. Therefore there is one solution per degree celcius in the range from <math>0</math> to <math>(\frac 59)(1000-32) + 1=(\frac 59)(968) + 1=538.8</math>, meaning there are <math>539</math> solutions.
-Let <math>c</math> be a degree celcius, and <math>f=(9/5)c+32</math> rounded to the nearest integer. <math>|f-((9/5)c+32)|\leq 1/2</math> <math>|(5/9)(f-32)-c|\leq 5/18</math> so it must round to <math>c</math> because <math>5/18<1/2</math>. Therefore there is one solution per degree celcius in the range from <math>0</math> to <math>(5/9)(1000-32)=(5/9)(968)=537.8</math>, meaning there are <math>538</math> solutions.
 == See also ==

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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