Difference between revisions of "2007 AIME I Problems/Problem 5"
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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 === | ||
+ | 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: | ||
+ | |||
+ | <math>F=32</math>: Works | ||
+ | |||
+ | <math>F=33</math>: Doesn't work | ||
+ | |||
+ | <math>F=34</math>: work | ||
+ | |||
+ | <math>F=35</math>: Doesn’t work | ||
+ | |||
+ | <math>F=36</math>: Works | ||
+ | |||
+ | <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 | ||
+ | |||
+ | === 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 20:30, 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
. 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 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.