2007 AIME 1 #5: Fahrenheit screws us again....

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The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ 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?

I have no idea how to start on this problem. I tried putting in some numbers to see if that got some stuff rolling, but that didn't work at all. Any ideas?

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0
On

The outcome of the first rounding step depends on the starting temperature's remainder modulo $9$, so it's reasonable to write the initial temperature as $9q+r$ for some $(q,r)$ and see what happens for different values of $r$. For example, we have:

$$\frac59(9q+0 - 32) = 5q - \frac{160}{9} \leadsto 5q - 18$$

where I denote by $\leadsto$ the result of rounding. Then we can do the reverse calculation:

$$\frac95(5q-18) + 32 = 9q - 0.6 \leadsto 9q - 1.$$

So we see that when $r=0$, we don't get the same number back. (If this is enough of a hint, stop reading here and try to solve the problem on your own.)

We can do the same calculation for other values of $r$. For $r=1$, we have

$$(9q + 1) ^\circ F = \left(5q - \frac{155}{9}\right)^\circ C \leadsto (5q-17)^\circ C = (9q + 1.2)^\circ F \leadsto (9q+1)^\circ F$$ so we do get the original number back.

Given 3 hours to do 15 problems, it is not unreasonable to take the time to just continue on in this way for $r=2,3,4,5,6,7,8$. We conclude that $r=1,3,5,7,8$ give back the original number, and the rest don't.

Starting at a temperature between $32$ and $1000$ means starting at $9\cdot 3 + 5$ and ending at $9\cdot 111 + 1$. So:

  • $r=1$ occurs with $q = 4,5,\dots, 111$, a total of $108$ times.
  • $r=3$ occurs with $q = 4,5,\dots, 110$, a total of $107$ times.
  • $r=5$ occurs with $q = 3,4,\dots, 110$, a total of $108$ times.
  • $r=7$ occurs with $q = 3,4,\dots, 110$, a total of $108$ times.
  • $r=8$ occurs with $q = 3,4,\dots, 110$, a total of $108$ times.

The final answer is therefore $539$.

0
On

You could start by eliminating the integers that cannot be solutions to $F \approx \frac{9}{5}C + 32$

Note,
$C \equiv 0 \pmod 5 \implies F \equiv 5 \pmod 9$
$C \equiv 1 \pmod 5 \implies F \equiv 7 \pmod 9$
$C \equiv 2 \pmod 5 \implies F \equiv 0 \pmod 9$
$C \equiv 3 \pmod 5 \implies F \equiv 1 \pmod 9$
$C \equiv 4 \pmod 5 \implies F \equiv 3 \pmod 9$

Now we check to make sure valid starting Farenheit temperatures map to the correct Celsius modular class.

$F = 9k + 5 \implies C = 5k' \cong 0 \pmod 5$
$F = 9k + 7 \implies C = 5k' + \frac{10}{9} \cong 1 \pmod 5$
$F = 9k + 0 \implies C = 5k' + \frac{20}{9} \cong 2 \pmod 5$
$F = 9k + 1 \implies C = 5k' + \frac{25}{9} \cong 3 \pmod 5$
$F = 9k + 3 \implies C = 5k' + \frac{35}{9} \cong 4 \pmod 5$

So we're all good. @misha above already counted them for you.