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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Function and symmetry | AIME I, 1984 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry. You may use sequential hints.

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

Key Concepts


Functions

Symmetry

Number Theory

Check the Answer


But try the problem first…

Answer: is 401.

Source
Suggested Reading

AIME I, 1984, Question 12

Elementary Number Theory by David Burton

Try with Hints


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.

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