Divizibilitate (O.J.)

Marcelina Popa · Post by **Marcelina Popa** » Sun Nov 02, 2008 2:06 pm

1. Stiind ca numarul \( \overline{abc} \) este divizibil prin \( a+b+c \), aratati ca si numarul \( \overline{(b+c)(a+c)(a+b)} \) este divizibil prin \( a+b+c \).

2. Aratati ca numarul
\( N=7^0+7^1+7^2+7^3+...+7^{1999} \)
se divide cu \( 400 \).

miruna.lazar · Post by **miruna.lazar** » Mon Nov 03, 2008 7:16 pm

2 . Este foarte simplu.

Grupam cei 2000 de termeni in grupe de cate 5. Avem :

\( N = ( 7^0 + 7^1 + 7^2 + 7^3 + 7^4 ) + ( 7^5 + 7^6 + 7^7 + 7^8 + 7^9 ) + ... + ( 7^{1995} + 7^ {1996} + 7^{1997} + 7^{1998} + 7^{1999} )

N= 7^0 ( 1 + 7 + 7^2 + 7^3 ) + 7^5 ( 1 + 7 + 7^2 + 7^3 ) +...+ 7^{1995} ( 1 + 7 + 7^2 + 7^3 )

N = 7^0 \cdot 400 + 7^5 \cdot 400 + ... + 7^{1995} \cdot 400 \)

\( N = 400 ( 7^0 + 7^5 + ... + 7^{1995} ) \)

Deci , => \( N \) este divizibil cu 400

Ma mai gandesc la 1

Marcelina Popa · Post by **Marcelina Popa** » Mon Nov 03, 2008 10:32 pm

Aoleu, Miruna, cat de cascata esti! Ai pornit cu grupe de cate 5, apoi ai dat factor comun in grupe de cate 4... Ideea e buna, dar trebuie revazute calculele.

Dorobantu Razvan · Post by **Dorobantu Razvan** » Tue Nov 04, 2008 10:29 am

Cum se pune bara deasupra? Am uitat

miruna.lazar · Post by **miruna.lazar** » Wed Nov 05, 2008 8:01 pm

\overline

miruna.lazar · Post by **miruna.lazar** » Wed Nov 05, 2008 8:15 pm

Da , am gresit . Trebuia sa fac in grupe de cate 4 .

\( N = 7^0 (1 + 7 + 7^1 + 7^2 + 7^3 ) + 7^4 ( 1 + 7 + 7^1 + 7^2 + 7^3 ) + ... + 7^{1994} ( 1 + 7 + 7^1 + 7^2 + 7^3 ) = N = 7^0 \cdot 400 + 7^4 \cdot 400 + ... + 1994 \cdot 400 = N = 400 \cdot ( 7^0 + 7^4 +...+7^{1994} ) \)

miruna.lazar · Post by **miruna.lazar** » Wed Nov 05, 2008 9:28 pm

E bine acum ?

Marcelina Popa · Post by **Marcelina Popa** » Wed Nov 05, 2008 11:20 pm

Ah... nu! (uite asa o sa ajung eu Thales cu mesaje bisilabice

)

miruna.lazar · Post by **miruna.lazar** » Thu Nov 06, 2008 8:35 pm

Am intrat complet in ceata. Sper ca acum am iesit din ea.

Deci , din nou : \( N = 7^0 ( 1 + 7 + 7^2 + 7^3 ) + 7^4 ( 1 + 7 + 7^2 + 7^3 ) +...7^{1996} ( 1 + 7 + 7^2 + 7^3 ) = N = 7^0 \cdot 400 + 7^4 \cdot 400 + ...+7^{1996} \cdot 400 = N = 400 ( 7^0 + 7^4 +...+ 7^{1996} ) \)

miruna.lazar · Post by **miruna.lazar** » Thu Nov 06, 2008 10:38 pm

Bun . Acum presupun ca am lucrat mai bine , deci trec la problema 1

\( \overline {abc} = 100a + 10b + c \vdots ( a +b +c ) \)

\( \overline { (b+c ) (a+c)(a+b) } = ( b + c ) 100 + ( a +c )10 + (a+b) \)

\( = 100b + 100c + 10a +10c + a + b = \)
\( = 11a + 101b + 110c \)
\( = ( 11a + 11b + 11 c ) + ( 90b+99c ) \)
\( = 11 ( a+ b+ c ) + ( 90b + 99c ) \)-> Notam relatia cu *

1) a + b +c | 100 + 10b + c
2) a + b + c | a+ b + c => a + b +c | 100 ( a + b+ c )

=> a+ b + c |100a + 100b + 100c
(1) = a +b +c | 100a+10b+c

=> a+b+c | 100a+100b+100c - 100a - 10b- c
=> a + b + c | 90b + 99c -> Notam relatia cu **

Cum a+b+c |11 ( a+b+c)
a+b+c | 90b + 99c =>

a + b+c | 11 ( a+b+c ) + 90b + 99c = \( \overline { (b+c)(a+c)(a+b) \)

alex2008 · Post by **alex2008** » Sat Nov 08, 2008 8:14 pm

La 1

\( \overline{(b+c)(a+c)(a+b)}=\overline{bca}+\overline{cab} \)

Stim ca \( \overline{abc}+\overline{bca}+\overline{cab}=111(a+b+c) \)

Dar \( a+b+c \) divide \( \overline{abc} \) rezulta \( a+b+c \)divide \( \overline{(b+c)(a+c)(a+b)} \)

Marcelina Popa · Post by **Marcelina Popa** » Sat Nov 08, 2008 8:26 pm

Acum e foarte bine, Miruna. Bravo

.

A ta, Alex, e frumoasa, dar cam concisa

.