Difference between revisions of "2007 AIME I Problems/Problem 5"
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== Solution 2 == | == 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 == | ||
| + | A full solution: | ||
| + | |||
| + | Let <math>c</math> be a degree celcius, and <math>f=(9/5)c+32</math> rounded to the nearest integer. <math>|f-((9/5)c+32)|\leq 1/2</math> <math>|(5/9)(f-32)-c|\leq 5/18</math> so it must round to <math>c</math> because <math>5/18<1/2</math>. Therefore there is one solution per degree celcius in the range from <math>0</math> to <math>(5/9)(1000-32)=(5/9)(968)=537.8</math>, meaning there are <math>538</math> solutions. | ||
== See also == | == See also == | ||
Revision as of 19:31, 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
Examine 
modulo 9.
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 | ||
