INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 
Bose Olympiad Project Round is Live now. Learn More 

June 6, 2020

Digits and Numbers | AIME I, 2012 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on digits and numbers.

Digits and numbers - AIME I, 2012


Let S be set of all perfect squares whose rightmost three digits in base 10 are 256. T be set of numbers of form \(\frac{x-256}{1000}\) where x is in S, find remainder when 10th smallest element of T is divided by 1000.

  • is 107
  • is 170
  • is 840
  • cannot be determined from the given information

Key Concepts


Digits

Algebra

Numbers

Check the Answer


Answer: is 170.

AIME I, 2012, Question 10

Elementary Number Theory by David Burton

Try with Hints


First hint

x belongs to S so perfect square, Let x=\(y^{2}\), here \(y^{2}\)=1000a+256 \(y^{2}\) element in S then RHS being even y=2\(y_1\) then \(y_1^{2}=250a+64\) again RHS being even \(y_1=2y_2\) then \(y_2^{2}\)=125\(\frac{a}{2}\)+16 then both sides being integer a=2\(a_1\) then \(y_2^{2}=125a_1+16\)

Second Hint

\(y_2^{2}-16=125a_1\) then \((y_2-4)(y_2+4)=125a_1\)

or, one of \((y_2+4)\) and \((y_2-4)\) contains a non negative multiple of 125 then listing smallest possible values of \(y_2\)

or, \(y_2+4=125\) gives \(y_2=121\) or, \(y_2-4=125\) gives \(y_2=129\) and so on

or, \(y_2=4,121,129,upto ,621\) tenth term 621

Final Step

\(y=4y_2\)=2484 then \(\frac{2483^{2}-256}{1000}\)=170.

Subscribe to Cheenta at Youtube


Leave a Reply

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

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com