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 .

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- August 5th 2007, 06:04 PM #1

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- August 6th 2007, 10:55 AM #2

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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, 08:13 PM #3

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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.

- August 13th 2007, 03:26 AM #4

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- June 17th 2009, 02:33 PM #5
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 . Counting carefully we get zeroes.