# Math Help - Problem 33

1. ## Problem 33

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 $-2007\leq x\leq 2007$.

2. 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].

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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.

4. Oops I forgot to count zero

5. The operation "reflection about 2" is the map $s: x \mapsto 4-x$

The operation "reflection about 7" is the map $t: x \mapsto 14-x$

Starting with $x=0$ we can obtain the zeroes of $f$ by repeatedly applying $s,t$ in any combination. We have

$t \circ s (x)= 14-(4-x)=10+x$

Moreover $s^2=t^2=1$, the identity map, so we can consider only alternating sequences of $s,t$. If we compose $(t \circ s)$ with itself n times then $(t \circ s)^n (x) = 10+(10+(10+...+(10+x)...)= 10n+x$ (where the exponent means composition - careful!)

So now we have three possible types of different sequences of composition:
$(t \circ s)^n\: (x) = 10n+x$
$s\circ (t \circ s)^n\: (x) = 4-(10n+x)$
$(t \circ s)^n \circ s\: (x) = 10n+(4-x)$

Since $x=0$ is our starting point, the zeroes of $f$ are of the form $10n, 10n+4, -10n+4$ for all $n \in \mathbb N$. It is easy to see that they are almost in bijection with the multiples of 5 on the interval $[-2007,2007]$. Counting carefully we get $2\Big\lfloor\frac{2007}{5}\Big\rfloor+2=804$ zeroes.