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== Problem == | == Problem == | ||
The formula for converting a Fahrenheit temperature <math>F</math> to the corresponding Celsius temperature <math>C</math> is <math>C = \frac{5}{9}(F-32).</math> An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest [[integer]]. | The formula for converting a Fahrenheit temperature <math>F</math> to the corresponding Celsius temperature <math>C</math> is <math>C = \frac{5}{9}(F-32).</math> An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest [[integer]]. | ||
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For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature? | For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature? | ||
| − | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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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. | 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. | + | 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 = \boxed{539}</math> as the solution. |
=== Solution 2 === | === Solution 2 === | ||
| − | Notice that <math>\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k</math> holds if <math>k=\left[ \frac{9}{5}x\right]</math> for some <math>x</math>. | + | Notice that <math>\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k</math> holds if <math>k=\left[ \frac{9}{5}x\right]</math> for some integer <math>x</math>. |
| − | Thus, after translating from <math>F\to F-32</math> we want count how many values of <math>x</math> there are such that <math>k=\left[ \frac{9}{5}x\right]</math> is an integer from <math>0</math> to <math>968</math>. This value is computed as <math>\left[968*\frac{5}{9}\right]+1</math>, adding in the extra solution corresponding to <math>0</math>. | + | Thus, after translating from <math>F\to F-32</math> we want count how many values of <math>x</math> there are such that <math>k=\left[ \frac{9}{5}x\right]</math> is an integer from <math>0</math> to <math>968</math>. This value is computed as <math>\left[968*\frac{5}{9}\right]+1 = \boxed{539}</math>, adding in the extra solution corresponding to <math>0</math>. |
| + | |||
| + | ==== Note ==== | ||
| + | Proof that <math>\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k</math> iff <math>k=\left[ \frac{9}{5}x\right]</math> for some integer <math>x</math>: | ||
| + | |||
| + | First assume that <math>k</math> cannot be written in the form <math>k=\left[ \frac{9}{5}x\right]</math> for any integer <math>x</math>. Let <math>z = \left[ \frac{5}{9}k\right]</math>. Our equation simplifies to <math>k = \left[ \frac{9}{5}z\right]</math>. However, this equation is not possible, as we defined <math>k</math> such that it could not be written in this form. Therefore, if <math>k \neq \left[ \frac{9}{5}x\right]</math>, then <math>\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] \neq k</math>. | ||
| + | |||
| + | Now we will prove that if <math>k = \left[ \frac{9}{5}x\right]</math>, <math>\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k</math>. We realize that because of the 5 in the denominator of <math>\left[ \frac{9}{5}x \right]</math>, <math>\left[ \frac{9}{5}x \right]</math> will be at most <math>\frac{2}{5}</math> away from <math>\frac{9}{5}x</math>. Let <math>z = \left[ \frac{9}{5}x \right]- \frac{9}{5}x</math>, meaning that <math>-\frac{2}{5} \leq z \leq \frac{2}{5}</math>. Now we substitute this into our equation: | ||
| + | |||
| + | <cmath>\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]</cmath>. | ||
| + | |||
| + | Now we use the fact that <math>-\frac{2}{5} \leq z \leq \frac{2}{5}</math> | ||
| + | |||
| + | <cmath>\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]</cmath> | ||
| + | |||
| + | <cmath>\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]</cmath> | ||
| + | |||
| + | Hence <math>\left[ \frac{9}{5} \left[ \frac{5}{9}k \right] \right] = \left[ \frac{9}{5}x \right] = k</math> and we are done. | ||
| + | |||
| + | - mako17 | ||
=== Solution 3 === | === Solution 3 === | ||
| − | Let <math>c</math> be a degree Celsius, and <math>f=\frac 95c+32</math> rounded to the nearest integer. Since <math>f</math> was rounded to the nearest integer we have <math>|f-((\frac 95)c+32)|\leq 1/2</math>, which is equivalent to <math>|(\frac 59)(f-32)-c|\leq \frac 5{18}</math> if we multiply by <math>5/9</math>. Therefore, it must round to <math>c</math> because <math>\frac 5{18}<\frac 12</math> so <math>c</math> is the closest integer. Therefore there is one solution per degree | + | Let <math>c</math> be a degree Celsius, and <math>f=\frac 95c+32</math> rounded to the nearest integer. Since <math>f</math> was rounded to the nearest integer we have <math>|f-((\frac 95)c+32)|\leq 1/2</math>, which is equivalent to <math>|(\frac 59)(f-32)-c|\leq \frac 5{18}</math> if we multiply by <math>5/9</math>. Therefore, it must round to <math>c</math> because <math>\frac 5{18}<\frac 12</math> so <math>c</math> is the closest integer. Therefore there is one solution per degree Celsius 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. |
=== Solution 4 === | === Solution 4 === | ||
Start listing out values for <math>F</math> and their corresponding values of <math>C</math>. You will soon find that every 9 values starting from <math>F</math> = 32, there is a pattern: | Start listing out values for <math>F</math> and their corresponding values of <math>C</math>. You will soon find that every 9 values starting from <math>F</math> = 32, there is a pattern: | ||
| − | Works | + | |
| − | + | <math>F=32</math>: Works | |
| − | + | ||
| − | Doesn’t work | + | <math>F=33</math>: Doesn't work |
| − | Works | + | |
| − | Works | + | <math>F=34</math>: work |
| − | Doesn’t work | + | |
| − | Works | + | <math>F=35</math>: Doesn’t work |
| − | Doesn’t work | + | |
| − | Works | + | <math>F=36</math>: 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 <math>107\cdot5 = 535</math> values of <math>F</math> 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 <math>535 + 4 = 539</math> solutions that work. | + | <math>F=37</math>: Works |
| + | |||
| + | <math>F=38</math>: Doesn’t work | ||
| + | |||
| + | <math>F=39</math>: Works | ||
| + | |||
| + | <math>F=40</math>: Doesn’t work | ||
| + | |||
| + | <math>F=41</math>: Works | ||
| + | |||
| + | There are <math>969</math> numbers between <math>32</math> and <math>1000</math>, inclusive. This is <math>107</math> sets of <math>9</math>, plus <math>6</math> extra numbers at the end. In each set of <math>9</math>, there are <math>5</math> “Works,” so we have <math>107\cdot5 = 535</math> values of <math>F</math> that work. | ||
| + | |||
| + | Now we must add the <math>6</math> extra numbers. The number of “Works” in the first <math>6</math> terms of the pattern is <math>4</math>, so our final answer is <math>535 + 4 = 539</math> solutions that work. | ||
| + | |||
Submitted by warriorcats | Submitted by warriorcats | ||
| + | |||
| + | === Solution 5(similar to solution 3 but faster solution if you have no time) === | ||
| + | Notice that every <math>C</math> value corresponds to exactly one <math>F</math> value but multiple <math>F</math> values can correspond to a <math>C</math> value. Thus, the smallest <math>C</math> value is <math>0</math> and the largest <math>C</math> value is <math>538</math> yielding <math>\boxed{539}</math> solutions. | ||
| + | |||
| + | -alanisawesome2018 | ||
== See also == | == See also == | ||
Latest revision as of 19:30, 4 August 2024
Contents
Problem
The formula for converting a Fahrenheit temperature 
to the corresponding Celsius temperature 
is 
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 
modulo 9.
- If
, then we can define 
. This shows that ![$F = \left[\frac{9}{5}\left[\frac{5}{9}(F-32)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x) + 32\right] \Longrightarrow F = 9x + 32$](//latex.artofproblemsolving.com/a/b/1/ab17851397d90651ccc4c0952de339d52b5efb3d.png)
. This case works. - If
, then we can define 
. This shows that ![$F = \left[\frac{9}{5}\left[\frac{5}{9}(F-32)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + 1) + 32\right] \Longrightarrow$](//latex.artofproblemsolving.com/0/5/8/0587851c2b314c48ed1acaf27347592719c2efda.png)
![$F = \left[9x + \frac{9}{5}+ 32 \right] \Longrightarrow F = 9x + 34$](//latex.artofproblemsolving.com/2/5/2/2521f37b106bfe5b2bac50d2540db93f0b95eb44.png)
. So this case doesn't work.
, then we can define 
. This shows that ![$F = \left[\frac{9}{5}\left[\frac{5}{9}(F-32)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x) + 32\right] \Longrightarrow F = 9x + 32$](http://latex.artofproblemsolving.com/a/b/1/ab17851397d90651ccc4c0952de339d52b5efb3d.png)
. This case works.
, then we can define 
. This shows that ![$F = \left[\frac{9}{5}\left[\frac{5}{9}(F-32)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + 1) + 32\right] \Longrightarrow$](http://latex.artofproblemsolving.com/0/5/8/0587851c2b314c48ed1acaf27347592719c2efda.png)
![$F = \left[9x + \frac{9}{5}+ 32 \right] \Longrightarrow F = 9x + 34$](http://latex.artofproblemsolving.com/2/5/2/2521f37b106bfe5b2bac50d2540db93f0b95eb44.png)
. So this case doesn't work.Generalizing this, we define that 
. Thus, ![$F = \left[\frac{9}{5}\left[\frac{5}{9}(9x + k)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + \left[\frac{5}{9}k\right]) + 32\right] \Longrightarrow F = \left[\frac{9}{5} \left[\frac{5}{9}k \right] \right] + 9x + 32$](http://latex.artofproblemsolving.com/9/f/6/9f64549ba35afaa22b01672a8da1b8fc26ee1b20.png)
. We need to find all values 
that ![$\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$](http://latex.artofproblemsolving.com/2/2/8/228d80bbd0cf97b9a3e2caf144c365bd2072c5f8.png)
. Testing every value of 
shows that 
, so 
of every 
values of work.
There are 
cycles of , giving 
numbers that work. Of the remaining 
numbers from 
onwards, 
work, giving us 
as the solution.
Solution 2
Notice that holds if ![$k=\left[ \frac{9}{5}x\right]$](http://latex.artofproblemsolving.com/3/c/a/3ca0c454d5e3beb0bbef5becedd9159c8a37ff04.png)
for some integer 
.
Thus, after translating from 
we want count how many values of there are such that is an integer from 
to 
. This value is computed as ![$\left[968*\frac{5}{9}\right]+1 = \boxed{539}$](http://latex.artofproblemsolving.com/f/b/c/fbce32cbf10d05488a495bd927f86bb4cf3b34f3.png)
, adding in the extra solution corresponding to .
Note
Proof that iff for some integer :
First assume that cannot be written in the form for any integer . Let ![$z = \left[ \frac{5}{9}k\right]$](http://latex.artofproblemsolving.com/9/0/a/90a475e726713833572129ace17d4e34787fca63.png)
. Our equation simplifies to ![$k = \left[ \frac{9}{5}z\right]$](http://latex.artofproblemsolving.com/2/c/6/2c62699135069d62cdcb61237a59e5719b4ef134.png)
. However, this equation is not possible, as we defined such that it could not be written in this form. Therefore, if ![$k \neq \left[ \frac{9}{5}x\right]$](http://latex.artofproblemsolving.com/b/9/0/b90b7e706392f419150a60f8524ecd40162cf183.png)
, then ![$\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] \neq k$](http://latex.artofproblemsolving.com/0/c/7/0c74fd7d675d622703a408a14efdd6d90b1a792d.png)
.
Now we will prove that if ![$k = \left[ \frac{9}{5}x\right]$](http://latex.artofproblemsolving.com/4/9/3/4933bb184fc458a66b7f0e302ec9c3dfeac7f0c7.png)
, . We realize that because of the 5 in the denominator of ![$\left[ \frac{9}{5}x \right]$](http://latex.artofproblemsolving.com/0/5/1/051ff6792a6f5280bb63a7e62954694689b43b8b.png)
, will be at most 
away from 
. Let ![$z = \left[ \frac{9}{5}x \right]- \frac{9}{5}x$](http://latex.artofproblemsolving.com/3/3/a/33a95ad7cfaa182dbb9ee2c1a0681035e2e40ba3.png)
, meaning that 
. Now we substitute this into our equation:
![\[\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9} \left[ \frac{9}{5}x\right] \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9} (\frac{9}{5}x + z) \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9} (\frac{9}{5}x + z) \right] \right] = \left[ \frac{9}{5} \left[ x+ \frac{5}{9}z \right] \right]\]](http://latex.artofproblemsolving.com/3/d/e/3de1aaef4fe711237918a23f778c68485a1aa417.png)
.
Now we use the fact that
![\[\left[ \frac{9}{5} \left[ x - \frac{5}{9}(\frac{2}{5}) \right] \right] \leq \left[ \frac{9}{5} \left[ x + \frac{5}{9}(z) \right] \right] \leq \left[ \frac{9}{5} \left[ x + \frac{5}{9}(\frac{2}{5}) \right] \right]\]](http://latex.artofproblemsolving.com/0/9/3/0932c8c38c41af4d1a94b737b6bc165e0e873ad2.png)
![\[\left[ \frac{9}{5} x \right] \leq \left[ \frac{9}{5} \left[ x + \frac{5}{9}(z) \right] \right] = \left[ \frac{9}{5} \left[ \frac{5}{9}k \right] \right] \leq \left[ \frac{9}{5}x \right]\]](http://latex.artofproblemsolving.com/4/1/b/41bc062a3d89fb9de7fc67619a58bb140206e874.png)
Hence ![$\left[ \frac{9}{5} \left[ \frac{5}{9}k \right] \right] = \left[ \frac{9}{5}x \right] = k$](http://latex.artofproblemsolving.com/8/5/c/85c122b1d3d89309a2345e2cbcd3d2b0995bcded.png)
and we are done.
- mako17
Solution 3
Let 
be a degree Celsius, and 
rounded to the nearest integer. Since 
was rounded to the nearest integer we have 
, which is equivalent to 
if we multiply by 
. Therefore, it must round to because 
so is the closest integer. Therefore there is one solution per degree Celsius in the range from to 
, meaning there are 
solutions.
Solution 4
Start listing out values for and their corresponding values of . You will soon find that every 9 values starting from = 32, there is a pattern:
: Works
: Doesn't work
: work
: Doesn’t work
: Works
: Works
: Doesn’t work
: Works
: Doesn’t work
: Works
There are 
numbers between 
and 
, inclusive. This is 
sets of , plus extra numbers at the end. In each set of , there are “Works,” so we have 
values of that work.
Now we must add the extra numbers. The number of “Works” in the first terms of the pattern is 
, so our final answer is 
solutions that work.
Submitted by warriorcats
Solution 5(similar to solution 3 but faster solution if you have no time)
Notice that every value corresponds to exactly one value but multiple values can correspond to a value. Thus, the smallest value is and the largest value is 
yielding 
solutions.
-alanisawesome2018
See also
| 2007 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
