Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on digits and numbers.
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.
Digits
Algebra
Numbers
Answer: is 170.
AIME I, 2012, Question 10
Elementary Number Theory by David Burton
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 \(y_2^{2}=125a_1+16\)
\(y_2^{2}-16=125a_1\) 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
\(y=4y_2\)=2484 then \(\frac{2483^{2}-256}{1000}\)=170.
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