Problem 33

ThePerfectHacker · #1 $Old$ 08-05-2007, 08:04 PM

Let $f(x)$ be a function on the number line.
It is given that $f(2+x)=f(2-x) \mbox{ and }f(7+x)=f(7-x)$ and $0$ is a zero of $f(x)$ . What is the least number of zeros must $f(x)$ have on the interval $-2007\leq x\leq 2007$ .

mathisfun1 · #2 $Old$ 08-06-2007, 12:55 PM

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 $R_2$ is a reflection about x=2 and $R_7$ about x=7, then we obtain a sequence of zeros $R_2, \ R_7R_2, \ R_2R_7R_2, ...$ or $R_7, \ R_2R_7, \ R_7R_2R_7, ...$ . If $R_2$ is our first reflection, then we obtain the following sequence of zeros $a_0=0, \ a_1=4, \ a_2 = 10, \ a_3 = -6, \ a_4 = 20$ and in general $a_{2m}=10m, \ a_{2m-1} = 14-10m$ , giving us 200+202 = 402 zeros in the interval [-2007, 2007]. If $R_7$ is our first reflection, then we get the sequence $a_{2k}=-10k, \ a_{2k-1} = 10k+4$ , 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].

ThePerfectHacker · #3 $Old$ 08-12-2007, 10:13 PM

Your solution seems correct.
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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.

mathisfun1 · #4 $Old$ 08-13-2007, 05:26 AM

Oops I forgot to count zero