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]]. |
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? | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
Examine <math>F - 32</math> modulo 9. | Examine <math>F - 32</math> modulo 9. | ||
− | == Solution 2 == | + | |
+ | *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. | ||
+ | * 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | |||
+ | === Solution 2 === | ||
Hint. Consider the identity <math>Round(ax)=Round(xRound(a/Round(ax))</math> its something like that... | Hint. Consider the identity <math>Round(ax)=Round(xRound(a/Round(ax))</math> its something like that... | ||
== Solution 3 == | == Solution 3 == |
Revision as of 19:41, 15 March 2007
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
Hint. Consider the identity its something like that...
Solution 3
A full solution:
Let be a degree celcius, and
rounded to the nearest integer.
so it must round to
because
. Therefore there is one solution per degree celcius in the range from
to
, meaning there are
solutions.
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 |