2007 AIME I Problem 5

Below is the professionally curated solution for Problem 5 of the 2007 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AIME I solutions, or check the answer key.

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Concepts:floor and ceiling functionsmodular arithmeticsmall cases

Difficulty rating: 2560

5.

The formula for converting a Fahrenheit temperature FF to the corresponding Celsius temperature CC is C=59(F32).C = \frac{5}{9}(F - 32). An integer Fahrenheit temperature is converted to Celsius and rounded to the nearest integer; the resulting integer Celsius temperature is converted back to Fahrenheit and rounded to the nearest integer. For how many integer Fahrenheit temperatures TT with 32T100032 \le T \le 1000 does the original temperature equal the final temperature?

Solution:

Adding 99 to FF adds exactly 55 to 59(F32),\frac{5}{9}(F - 32), hence 55 to the rounded Celsius value, hence 99 to the final Fahrenheit value. So TT returns to itself if and only if T+9T + 9 does, and it suffices to check nine consecutive temperatures. Checking 3232 through 40:40: the final values are 32,34,34,36,36,37,37,39,39,32, 34, 34, 36, 36, 37, 37, 39, 39, so exactly the five temperatures 32,34,36,37,3932, 34, 36, 37, 39 survive.

The range from 3232 through 994994 contains 963=1079963 = 107 \cdot 9 integers, contributing 1075=535107 \cdot 5 = 535 survivors. The remaining 995,,1000995, \ldots, 1000 behave like 32,,37,32, \ldots, 37, of which 32,34,36,3732, 34, 36, 37 survive, adding 44 more.

The total is 535+4=539.535 + 4 = 539.

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