Function and symmetry | AIME I, 1984 | Question 12

Join Trial or Access Free Resources

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


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.

Subscribe to Cheenta at Youtube


More Posts

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

linkedin facebook pinterest youtube rss twitter instagram facebook-blank rss-blank linkedin-blank pinterest youtube twitter instagram