Olimpíadas ⇒ (OBM) Algarismo das unidades

[mclaratrajano]

Se multiplicarmos todos os naturais menores que 2011 que não são múltiplos de 5, qual será o algarismo das unidades do número obtido?

a) 2
b)4
c)6
d)7
e)8

Resposta

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c

Olá! Alguém me dá uma luz? Como posso resolver essa questão?

[rcompany]

[tex3]
\text{Seja }N\in\mathbb{N}^*\\
\prod_{k=1}^{5N}k=\prod_{k=0}^{N-1}\left(\prod_{i=1}^{5}(5k+i)\right)\\[12pt]
\begin{array}{rl}\text{e }A_{5N}=\dfrac{\displaystyle\prod_{k=1}^{N-1}k}{\displaystyle\prod_{k=1}^{N}5k}&=\displaystyle\prod_{k=0}^{N-1}\left(\prod_{i=1}^{4}(5k+i)\right)\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(5k+1)(5k+4)(5k+2)(5k+3)\right]\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(25k^2+25k+4)(25k^2+25k+6)\right]\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(25k^2+25k+5-1)(25k^2+25k+5+1)\right]\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(25k^2+25k+5)^2-1\right]\\
\end{array}\\
\forall k\in\mathbb{N}, \exists \beta_k\in\mathbb{N}\text{ tal que }(25k^2+25k+5)^2=\beta_k\cdot 10+5\\
\text{ e então }A_{5N}=\displaystyle\prod_{k=0}^{N-1}\left[\beta_k\cdot 10+4\right]\\
A_5=(25+25+5)^2=55^2-1=3024=302\times 10+4=\gamma_{1}\cdot 10+4\\[6pt]
\text{Supondo que para }p\in\mathbb{N},\ \exists \gamma_{2p+1}\!\in\!\mathbb{R}\text{ tal que }A_{5(2p+1)}=\gamma_{2p+1}\cdot 10+4\\[12pt]
\text{então }A_{5(2p+2)}=(\gamma_{2p+1}\cdot 10+4)(\beta_{2p+2}\cdot 10+4)=10\left[(\gamma_{2p+1}\beta_{2p+2}\cdot 10+4(\gamma_{2p+1}+\beta_{2p+2})+1\right]+6=\gamma_{2p+2}\cdot 10+6\\[12pt]
\text{e }A_{5(2p+3)}=(\gamma_{2p+2}\cdot 10+4)(\beta_{2p+3}\cdot 10+4)=10\left[(\gamma_{2p+2}\beta_{2p+3}\cdot 10+4(\gamma_{2p+2}+\beta_{2p+3})+2\right]+4=\gamma_{2p+3}\cdot 10+4\\[18pt]
\text{ e então existe uma sequencia }(\gamma_n)_{n\in\mathbb{N}}\text{ com valores em }\mathbb{R}\text{ tal que:}\\
\forall p\in\mathbb{N},\left\{\begin{array}{l}A_{5(2p)}=\gamma_{2p}\cdot 10+6\\A_{5(2p+1)}=\gamma_{2p+1}\cdot 10+4\end{array}\right.\\[18pt]
A_{2010}=A_{5\times 402}\text{ e }\exists \gamma_{402}\!\in\!\mathbb{R}\text{ tal que }A_{2010}=\gamma_{402}\cdot 10+6,\text{ ie.}\\[18pt]
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\text{Seja }N\in\mathbb{N}^*\\
\prod_{k=1}^{5N}k=\prod_{k=0}^{N-1}\left(\prod_{i=1}^{5}(5k+i)\right)\\[12pt]
\begin{array}{rl}\text{e }A_{5N}=\dfrac{\displaystyle\prod_{k=1}^{N-1}k}{\displaystyle\prod_{k=1}^{N}5k}&=\displaystyle\prod_{k=0}^{N-1}\left(\prod_{i=1}^{4}(5k+i)\right)\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(5k+1)(5k+4)(5k+2)(5k+3)\right]\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(25k^2+25k+4)(25k^2+25k+6)\right]\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(25k^2+25k+5-1)(25k^2+25k+5+1)\right]\\
&=\displaystyle\prod_{k=0}^{N-1}\left[(25k^2+25k+5)^2-1\right]\\
\end{array}\\
\forall k\in\mathbb{N}, \exists \beta_k\in\mathbb{N}\text{ tal que }(25k^2+25k+5)^2=\beta_k\cdot 10+5\\
\text{ e então }A_{5N}=\displaystyle\prod_{k=0}^{N-1}\left[\beta_k\cdot 10+4\right]\\
A_5=(25+25+5)^2=55^2-1=3024=302\times 10+4=\gamma_{1}\cdot 10+4\\[6pt]
\text{Supondo que para }p\in\mathbb{N},\ \exists \gamma_{2p+1}\!\in\!\mathbb{R}\text{ tal que }A_{5(2p+1)}=\gamma_{2p+1}\cdot 10+4\\[12pt]
\text{então }A_{5(2p+2)}=(\gamma_{2p+1}\cdot 10+4)(\beta_{2p+2}\cdot 10+4)=10\left[(\gamma_{2p+1}\beta_{2p+2}\cdot 10+4(\gamma_{2p+1}+\beta_{2p+2})+1\right]+6=\gamma_{2p+2}\cdot 10+6\\[12pt]
\text{e }A_{5(2p+3)}=(\gamma_{2p+2}\cdot 10+4)(\beta_{2p+3}\cdot 10+4)=10\left[(\gamma_{2p+2}\beta_{2p+3}\cdot 10+4(\gamma_{2p+2}+\beta_{2p+3})+2\right]+4=\gamma_{2p+3}\cdot 10+4\\[18pt]
\text{ e então existe uma sequencia }(\gamma_n)_{n\in\mathbb{N}}\text{ com valores em }\mathbb{R}\text{ tal que:}\\
\forall p\in\mathbb{N},\left\{\begin{array}{l}A_{5(2p)}=\gamma_{2p}\cdot 10+6\\A_{5(2p+1)}=\gamma_{2p+1}\cdot 10+4\end{array}\right.\\[18pt]
A_{2010}=A_{5\times 402}\text{ e }\exists \gamma_{402}\!\in\!\mathbb{R}\text{ tal que }A_{2010}=\gamma_{402}\cdot 10+6,\text{ ie.}\\[18pt]
[/tex3]

[tex3]
\boxed{\text{6 é o algarismo das unidades de }A_{2010}}\\[12pt]
\rule{15cm}{1px}\\[12pt]
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\boxed{\text{6 é o algarismo das unidades de }A_{2010}}\\[12pt]
\rule{15cm}{1px}\\[12pt]
[/tex3]