Difference between revisions of "2007 AIME I Problems/Problem 5"
m (→Solution 2) |
m (→Solution 1) |
||
Line 9: | Line 9: | ||
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{ | + | 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 = 539</math> as the solution. |
Revision as of 19:29, 11 July 2011
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
.
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
.
Solution 3
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 |