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Number theory Difficulty 4.7 AIME Prove it Ukraine

We know that at some natural $n$ the number $n^2 + 2008n$ written in decimal notation ends with 4. Find what digit is in the ten's place of the number.

Solution

It's clear that the number $2000n$ does not influence the answer, which implies that the sought digits will be the same for numbers $A = n^2 + 2008n$ and $B = n^2 + 8n^2$ . As number $(B+16)$ equals $(n+4)^2$ (being the square of the natural number) and ends in $0$ , this number should end in $00$ . Thus $B = \overline{X00}-16 = \overline{Y84}$ where $X, Y$ are some natural numbers. Therefore the last two digits of the number are $84$ .

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