Difference between revisions of "2007 AIME I Problems/Problem 5"
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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? | |
− | |||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
Examine <math>F - 32</math> modulo 9. | Examine <math>F - 32</math> modulo 9. | ||
− | *If <math> | + | *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. |
− | * If <math> | + | * 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. |
+ | |||
+ | 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 = \boxed{539}</math> as the solution. | ||
+ | |||
+ | === 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 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>. | ||
+ | |||
+ | ==== 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>. | ||
− | === Solution | + | Now we use the fact that <math>-\frac{2}{5} \leq z \leq \frac{2}{5}</math> |
− | + | ||
− | == Solution 3 == | + | <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 === | ||
+ | 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. | ||
+ | |||
+ | === 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 == | ||
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 08:22, 25 July 2024
Contents
How many of the numbers in the listare rounded up when rounded to the nearest thousandth?
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
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.