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Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry.

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

Functions

Symmetry

Number Theory

But try the problem first...

Answer: is 401.

Source

Suggested Reading

AIME I, 1984, Question 12

Elementary Number Theory by David Burton

First hint

by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots

Second Hint

here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)

Final Step

Then least number of roots as 401.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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