Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry.
Function and Symmetry – AIME I 1984
A function f is defined for all real numbers and satisfies f(2+x)=f(2-x) and f(7+x)=f(7-x) for all x. If x=0 is root for f(x)=0, find the least number of roots f(x) =0 must have in the interval \(-1000 \leq x\leq 1000\).
- is 107
- is 401
- is 840
- cannot be determined from the given information
Check the Answer
But try the problem first…
Answer: is 401.
AIME I, 1984, Question 12
Elementary Number Theory by David Burton
Try with Hints
by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots
here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)
Then least number of roots as 401.
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