(nu sunt numere naturale nenule! , sunt cifre)
\( a+d=x \)
\( b+c=x \)
\( \overline{abcd}=1000a+d+100b+10c=1000x+110x=1110x=> \)\( 1110 \)nu este divizibil=>x \( | 40 => x\in {1,2,4,8}=> \) \( a \) \( diferit \) \( de \) \( 0 \) \( =>a=1 \),\( d=0 \)
Search found 11 matches
- Fri Mar 21, 2008 9:44 pm
- Forum: Clasa a V-a
- Topic: Concurs "Teodor Topan" - problema 1
- Replies: 2
- Views: 862
- Fri Mar 21, 2008 9:39 pm
- Forum: Clasa a V-a
- Topic: Concurs "Teodor Topan" - problema 1
- Replies: 2
- Views: 862
- Fri Mar 21, 2008 7:51 pm
- Forum: Clasa a VI-a
- Topic: Patrat perfect !
- Replies: 5
- Views: 1098
- Fri Mar 21, 2008 7:50 pm
- Forum: Clasa a VI-a
- Topic: Patrat perfect !
- Replies: 5
- Views: 1098
- Fri Mar 21, 2008 7:48 pm
- Forum: Clasa a V-a
- Topic: 11 numere naturale distincte
- Replies: 1
- Views: 670
Sa presupunem ca numerele noastre sunt cele mai mici posibile , diferite de 0 ! => 1+2+3+...+11=(11+1)*11:2=12*11:2=6*11=66 => Un numar mai mare sau egal cu 8 trebuie marit cu 4 , deoarece , daca ar fi mai mic decat 8 atunci 1+4 ; 2+4 ; ... ; 7+4 . S-ar mai regasi ! => 420=2^{2}*3*5*7=4*3*5*7 => De ...
- Fri Mar 21, 2008 7:16 pm
- Forum: Clasa a VI-a
- Topic: Patrat perfect !
- Replies: 5
- Views: 1098
Patrat perfect !
Se da numarul de forma \( \bar {aabb} \) , patrat perfect , in baza 10.
- Fri Mar 21, 2008 2:48 pm
- Forum: Clasa a VI-a
- Topic: parageozilizm!!!
- Replies: 1
- Views: 643
- Fri Mar 21, 2008 2:12 pm
- Forum: Clasa a VI-a
- Topic: Concursul "Teodor Topan" - problema 1
- Replies: 3
- Views: 993
rezolvare
\( n=2^{2005}*(2-1)-2^{2004}-2^{2003}=2^{2004}*(2-1) - 2^{2003} = 2^{2003}*(2-1)=2^{2003} \)
\( \frac{n}{x}=\frac{8^{667}}{5} => (2^{3})^{667}*x = 2^{2003}*5 => 2^{2001}x=2^{2003}*5 => 2^{2003}:2^{2001}=x:5 => 4=x:5 => x=20 \)
\( \frac{n}{x}=\frac{8^{667}}{5} => (2^{3})^{667}*x = 2^{2003}*5 => 2^{2001}x=2^{2003}*5 => 2^{2003}:2^{2001}=x:5 => 4=x:5 => x=20 \)
- Fri Mar 21, 2008 2:09 pm
- Forum: Clasa a VI-a
- Topic: Concursul "Teodor Topan" - problema 1
- Replies: 3
- Views: 993
am gresit :D
11a+13=35 => 11a = 22 => a=2
- Fri Mar 21, 2008 2:08 pm
- Forum: Clasa a VI-a
- Topic: Concursul "Teodor Topan" - problema 1
- Replies: 3
- Views: 993
rezolvare
\( 11a + 13 = 35 => 11a=22 => a=2 \)