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, 2008

Number theory Difficulty 5.0 AIME Prove it Ukraine

We know that at some natural $n$ the number $n^3 + 2009n^2 + 27n$ written in decimal notation ends with digit 3. Find what digits are in the hundred's and ten's place of the number.

Solution

It's clear that the number $2000n^2$ does not influence the answer therefore the sought digits are the same for numbers $A = n^3 + 2009n^2 + 27n$ and $B = n^3 + 9n^2 + 27n$ .

As the number $(B + 27)$ equals $(n + 3)^3$ (being the cube of the natural number) and ends in 0, this number should end in 000. Thus $B = \overline{X000} - 27 = \overline{Y073}$ where $X, Y$ are some natural numbers. Therefore the last three digits of the number are 073.

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