Maraton de probleme de clasa a V-a - semestrul I

Luiza · Post by **Luiza** » Sun Dec 28, 2008 3:02 pm

Problema 18
Aratati ca numarul \( A=9^{2k}-7^{4k} \) este divizibil cu 10 .

mihai++ · Post by **mihai++** » Sun Dec 28, 2008 3:38 pm

\( A=-40B \)

miruna.lazar · Post by **miruna.lazar** » Mon Dec 29, 2008 7:06 pm

A= \( 9^{2k} - 7^{4k} \)

\( u(9^{2k}) = 1 \)

\( u(7^{4k}) = 1 \)

1-1 = 0 => \( \overline {............0 } \vdots 10 \)

Claudiu Mindrila · Post by **Claudiu Mindrila** » Mon Dec 29, 2008 8:35 pm

Determinati cifrele \( a,b,c,d \) pentru care \( c^{\overline{ab}}=\overline{abcd} \) si \( \frac{d}{c}=2 \).
Claudiu Mindrila, Suplimentul G.M.-B. 4/2008

mihai++ · Post by **mihai++** » Tue Dec 30, 2008 10:51 am

\( c^{\overline{ab}}=\overline{abcd}\Rightarrow c=2\Rightarrow d=4\Rightarrow \overline{abcd}=1024 \).

alex2008 · Post by **alex2008** » Sun Jan 04, 2009 1:02 am

Problema 20
Sa se determine cardinalul multimii : A={\( \overline{ab}\ |\ \overline{ab}=a\cdot b+a+b \) , unde a si b sunt cifre in baza 10 }

Quit · Post by **Quit** » Sun Jan 04, 2009 8:36 pm

\( 10a+b=ab+a+b \Leftrightarrow 9a=ab \Leftrightarrow b=9 \Leftrightarrow A=\{19,29,39,49,59,69,79,89,99\} \Leftrightarrow \) Card \( A=9 \)

Quit · Post by **Quit** » Sun Jan 04, 2009 8:42 pm

Problema 21
Sa se arate ca suma \( 2001^p+2001^p+...+2001^p \) , care are \( 2000 \) de termeni , este divizibila cu \( 87\cdot 10^3 \) , unde p este un numar natural nenul .

Luiza · Post by **Luiza** » Mon Jan 05, 2009 11:08 am

Dam factor comun pe \( 2001^p \)
\( S=2001^p\cdot 2000 \Rightarrow 1000 / S \)
\( 87 / 2001\Rightarrow 87 \cdot 10^3 \) divide \( S \)

Luiza · Post by **Luiza** » Mon Jan 05, 2009 11:16 am

Problema 22
Impartind un numar natural a la 63 obtinem restul 34 . Ce rest obtinem daca impartim numarul a la 21 ?

Marius Mainea · Post by **Marius Mainea** » Mon Jan 05, 2009 11:28 am

Luiza wrote:Problema 22
Impartind un numar natural a la 63 obtinem restul 34 . Ce rest obtinem daca impartim numarul a la 21 ?

63 se imparte exact la 21 deci restul impartirii numarului la 21 este restul impartirii lui 34 la 21 adica 13.

alex2008 · Post by **alex2008** » Mon Jan 05, 2009 12:17 pm

Vreti sa spuneti restul impartirii lui 34 la 21 adica 13 .

Claudiu Mindrila · Post by **Claudiu Mindrila** » Mon Jan 05, 2009 9:28 pm

Putin mai dificila.
Problema. Determinati restul impartirii numarului \( 2008^{2008} \) la \( 49 \).

Quit · Post by **Quit** » Tue Jan 06, 2009 8:58 am

\( 2008^{2008}=2008^{2007}\cdot 2008=2008^{2007}\cdot(2009-1)=2009\cdot 2008^{2007}-2008^{2007}=2009\cdot 2008^{2007}-2008^{2006}\cdot 2008=2009\cdot 2008^{2007}-2008^{2006}(2009-1)= \)
\( =2009\cdot 2008^{2007}-2009\cdot 2008^{2006}-2008^{2006}=...=2009\cdot 2008^{2007}-2009\cdot 2008^{2006}-2009\cdot 2008^{2005}-...-2008= \)

\( =2009\cdot 2008^{2007}-2009\cdot 2008^{2006}-2009\cdot 2008^{2005}-...-2009+1=2009(2008^{2007}-2008^{2006}-...-1)+1 \)
Stiind ca 2009 este divizibil cu 49 rezulta ca restul este 1 .

Quit · Post by **Quit** » Wed Jan 07, 2009 10:01 pm

Problema 24

Sa se afle \( n\in\mathb{N} \) din egalitatea :
\( n^2+n+1000^2=1+2+3+...+1999 \)

Luiza · Post by **Luiza** » Sun Jan 11, 2009 7:13 pm

\( n^2+n+1000^2=\frac{2000\cdot 1999}{2} \Leftrightarrow n^2+n =1000 \cdot 999 \Leftrightarrow n(n+1)=1000\cdot 999 \Leftrightarrow n=999 \) .

Luiza · Post by **Luiza** » Sun Jan 11, 2009 7:31 pm

Problema 25
Aratati ca numarul \( 5^{n+1} \) se poate scrie ca suma de cinci numere consecutive , oricare \( n\in \mathb{N}* \) .