Difference between revisions of "2007 AIME I Problems/Problem 5"
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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 === | ||
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=== 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 | ||
Latest revision as of 15:30, 3 May 2022
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?
Contents
Solution
Solution 1
Examine modulo 9.
- If , then we can define . This shows that . This case works.
- If , then we can define . This shows that . So this case doesn't work.
Generalizing this, we define that . Thus, . We need to find all values that . 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 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 , 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 . Our equation simplifies to . However, this equation is not possible, as we defined such that it could not be written in this form. Therefore, if , then .
Now we will prove that if , . We realize that because of the 5 in the denominator of , will be at most away from . Let , meaning that . Now we substitute this into our equation:
.
Now we use the fact that
Hence 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 celcius 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
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 |
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