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

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== Problem ==
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__TOC__
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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?
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How many of the numbers in the list<cmath>\left\{25.34816, \;\; 84.3695, \;\; 2.54527\cdot 10, \;\; 894.54332, \;\; \frac{234.572}{100}, \;\; \frac{162}{1000}\right\}</cmath>are rounded up when rounded to the nearest thousandth?
  
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__TOC__
 
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
 
Examine <math>F - 32</math> modulo 9.
 
Examine <math>F - 32</math> modulo 9.
  
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*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.
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*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.
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* 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 F = \left[9x + \frac{9}{5}+ 32 \right] \Longrightarrow F = 9x + 34</math>. So this case doesn't work.  
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* 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.  
  
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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{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.
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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.
  
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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.
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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 ===
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Hint. Consider the identity <math>Round(ax)=Round(xRound(a/Round(ax))</math> its something like that...
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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 = \boxed{539}</math>, adding in the extra solution corresponding to <math>0</math>.
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 +
==== Note ====
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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:
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 +
<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>
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 +
<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.
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 +
- mako17
  
 
=== Solution 3 ===
 
=== Solution 3 ===
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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.
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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 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.
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 +
=== Solution 4 ===
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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:
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<math>F=32</math>: Works
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<math>F=33</math>: Doesn't work
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<math>F=34</math>: work
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<math>F=35</math>: Doesn’t work
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<math>F=36</math>: Works
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<math>F=37</math>: Works
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<math>F=38</math>: Doesn’t work
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<math>F=39</math>: Works
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<math>F=40</math>: Doesn’t work
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 +
<math>F=41</math>: Works
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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.
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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.
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 +
Submitted by warriorcats
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 +
=== 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.
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 +
-alanisawesome2018
  
 
== See also ==
 
== See also ==
Line 26: Line 79:
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 08:22, 25 July 2024


How many of the numbers in the list\[\left\{25.34816, \;\; 84.3695, \;\; 2.54527\cdot 10, \;\; 894.54332, \;\; \frac{234.572}{100}, \;\; \frac{162}{1000}\right\}\]are rounded up when rounded to the nearest thousandth?

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 $535 + 4 = \boxed{539}$ 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 $0$.

Note

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

Solution 5(similar to solution 3 but faster solution if you have no time)

Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $\boxed{539}$ solutions.

-alanisawesome2018

See also

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

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