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. <math>|f-((\frac 95)c+32)|\leq 1/2</math> <math>|(\frac 59)(f-32)-c|\leq \frac 5{18}</math> so it must round to <math>c</math> because <math>\frac 5{18}<\frac 12</math>. 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. <math>|f-((\frac 95)c+32)|\leq 1/2</math> and <math>|(\frac 59)(f-32)-c|\leq \frac 5{18}</math> so it must round to <math>c</math> because <math>\frac 5{18}<\frac 12</math>. 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. |
== See also == | == See also == |
Revision as of 01:07, 27 June 2018
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 Celsius, and
rounded to the nearest integer.
and
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 |
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