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August 5th, 2007, 08:04 PM
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| | Problem 33 Let  be a function on the number line.
It is given that  and  is a zero of  . What is the least number of zeros must  have on the interval  . | 
August 6th, 2007, 12:55 PM
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| | The function is symmetric about x=2 and x=7, so all necessary zero are found by reflecting x=0 repeatedly about x=2 and x=7; if  is a reflection about x=2 and  about x=7, then we obtain a sequence of zeros  or  . If  is our first reflection, then we obtain the following sequence of zeros  and in general  , giving us 200+202 = 402 zeros in the interval [-2007, 2007]. If  is our first reflection, then we get the sequence  , which yields another 402 zeros. A quick check shows that the two sequences have no terms in common. Hence at least 804 zeros total in [-2007, 2007]. | 
August 12th, 2007, 10:13 PM
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| | Your solution seems correct.
----
This was a slighy modified problem from an AIME question.
We know f(0) is a zero.
Thus,
f(4) is a zero.
Thus,
f(10) is a zero.
Thus,
f(-6) is a zero.
Thus,
f(20) is a zero.
Thus,
f(-16) is a zero.
Thus,
f(30) is a zero.
....
The pattern is clear.
Similarly we get that,
f(14) is a zero.
Thus,
f(-10) is a zero.
Thus,
f(24) is a zero.
Thus,
f(-20) is a zero.
...
The pattern is clear.
Look at chart below.
RED: 4,-6,-16,...
BLUE: 14,24,34,...
GREEN: 10,20,30,...
GREY: -10,-20,-30,...
These are all arithmetic sequences.
Now we count how many are in this inteveral + 1 (because we never counted zero yet).
And also, these sequences are disjoint so we do not count the same thing twice. | 
August 13th, 2007, 05:26 AM
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| | Oops I forgot to count zero | 
June 17th, 2009, 04:33 PM
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| | The operation "reflection about 2" is the map
The operation "reflection about 7" is the map
Starting with  we can obtain the zeroes of  by repeatedly applying  in any combination. We have
Moreover  , the identity map, so we can consider only alternating sequences of  . If we compose  with itself n times then  (where the exponent means composition - careful!)
So now we have three possible types of different sequences of composition:
Since  is our starting point, the zeroes of  are of the form  for all  . It is easy to see that they are almost in bijection with the multiples of 5 on the interval ![[-2007,2007] [-2007,2007]](http://www.mathhelpforum.com/math-help/latex2/img/38a5676ed02ad881dc96d036a77b4cef-1.gif) . Counting carefully we get  zeroes. | | Thread Tools | | | | Display Modes | Linear Mode |
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