ex. 1 ___ ____ ___ ___ 12x + 13x + x37 +4x2 = 1 249 x ∈ { 0; 1;2; 3; 4; 5; 6; 7; 8; 9}(1·100+2·10+x·1) +(1·100+3·10+x·1)+(x·100+3·10+7·1) + (4·100+x·10+ 2·1)=1 249
1·100+2·10+x·1 +1·100+3·10+x·1+x·100+3·10+7·1+ 4·100+x·10+ 2·1=1 249
100( 1+ 1+ x+ 4) +10( 2+ 3+ 3+ x) + 1( x+ x+ 7+ 2) = 1 249
100( 6+ x) +10(8+ x) + 1( x+ x+ 9) = 1·1 000+ 2·100+ 4·10+ 9·1
1
. pt. ord. 1 = U x+ x+ 9 = 9
x+ x = 9- 9
x+ x = 0 , x=x= 0 sau
x= 5 , 5+ 5 = 1
0
2.
pt. ord. 2 = Z 10(8+ x) = 4·10
10(8+ x)+ 1 Z . preluată = 4·10
80+ x·10+ 10 = 40
90 + x·10 = 40 o
bs. Z termenului > Z sumei Se impune 140. 90+ x·10 =140
x·10 = 140- 90
x·10 = 50 I:10
x = 5 3.
pt. ord. 3 = S 100( 6+ x) + 1 S preluata = 2·100
600 + 100·x+ 100 = 200
Obs. S termenului > S sumei Se impune 1 200. 700+ 100·x = 1 200
100·x = 1 200- 700
100·x = 500 I: 100
x = 5Concluzie: cifra de la ord. 1= U ord. 2= Z ord. 3 = S ____________ x = 5
R: 5
___ ___ ___ ___
probă: 12x+ 13x+ x37+ 4x2= 1 249
125+ 135+537+ 452 = 1 249
1 249= 1 249
ex. 2
x= ?
________________________________
____
_5·3ⁿ+ 2·3ⁿ⁺¹+ 3ⁿ_
= _3ab_ x- 1≠ 0 , x≠ +1
2·3ⁿ x- 1
___
3ab : 18 , cel mai
mare nr.
___
1.
Se
află 3ab.
___ ___
3ab : 18 , 3ab : 2
, b = {0; 2; 4; 6; 8}
__
3ab : 9 , ( 3+ a+ b) : 9 ___
pt. a= 9 ( 3+ 9+ b) : 9 3ab = 396
( 12+ b) :
9
pt. b= 6 ( 12+ 6) : 9
18 : 9
A
2. Se
află x.
____
_5·3ⁿ+ 2·3ⁿ⁺¹+ 3n_
= _3ab_
2·3ⁿ x- 1
_5·3ⁿ+ 2·3ⁿ⁺¹+ 3ⁿ_
= _396_ , x- 1 = 0 , x = + 1
2·3ⁿ x- 1
_5·3ⁿ+ 2·3ⁿ ·3+ 3ⁿ·1_
= _396_
2·3ⁿ x- 1
_3ⁿ·( 5+ 2·3 + 1)_ =
_396_
2·3ⁿ x- 1
_ ( 5+ 6+ 1)_ = _396_
2 x- 1
_ 12 = _396_
2 x- 1
6 = _396_
x- 1
6( x- 1) = 396 I: 6
x- 1 = 66
x = 66+ 1
x = 67
