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




Check the Answer

But try the problem first…

Answer: is 170.

Suggested Reading

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.

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