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# 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

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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