Difference between revisions of "2007 AIME I Problems/Problem 5"
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=== 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 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. | 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: | ||
+ | Works | ||
+ | Works | ||
+ | Doesn’t work | ||
+ | Doesn’t work | ||
+ | Works | ||
+ | Works | ||
+ | Doesn’t work | ||
+ | Works | ||
+ | Doesn’t work | ||
+ | Works | ||
+ | There are 969 numbers between 32 and 1000, inclusive. This is 107 sets of 9, plus 6 extra numbers at the end. In each set of 9, there are 5 “Works,” so we have <math>107\cdot5 = 535</math> values of <math>F</math> that work. | ||
+ | Now we must add the 6 extra numbers. The number of “Works” in the first 6 terms of the pattern is 4, so our final answer is <math>535 + 4 = 539</math> solutions that work. | ||
+ | Submitted by warriorcats | ||
== See also == | == See also == |
Revision as of 19:25, 21 February 2020
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?
Contents
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 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 Works Doesn’t work Doesn’t work Works Works Doesn’t work Works Doesn’t work Works There are 969 numbers between 32 and 1000, inclusive. This is 107 sets of 9, plus 6 extra numbers at the end. In each set of 9, there are 5 “Works,” so we have values of that work. Now we must add the 6 extra numbers. The number of “Works” in the first 6 terms of the pattern is 4, so our final answer is solutions that work. Submitted by warriorcats
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 |
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