Răspuns:
14. 3*2^2009 / 5*2^2007 =
= 3*4 / 5 =
= 12/5 = 2 rest 2
15. abcd : bcd = a+1 rest a+2
Teorema impartirii cu rest :
D = C*Î + R, R < Î
abcd = bcd(a + 1) + a + 2
1000a + bcd - bcd(a + 1) - a = 2
999a + bcd(1 - a - 1) = 2
999a - a*bcd = 2
a(999 - bcd) = 2
1) a = 1 =>
999 - bcd = 2
bcd = 997
Deci abcd = 1997
2) a = 2
999 - bcd = 1
bcd = 998
abcd = 2998
Concluzia : abcd € {1997; 2998}
16. N = 5(n + 1) + 6^n+2 + 1001^n+3 + 5
Trb sa demonstram ca N != K^2
Notam U(x) ultima cifra a lui x.
U(K^2) € {0, 1, 2, 9, 6, 5}
U(5(n+1)) € {0, 5}
U(6^n+2) = 6
U(1001^n+3) = 1
Si mai avem un 5.
1) U(N) = U(0+6+1+5) = U(12) = 2
2 !€ U(K^2)
2) U(N) = U(5+6+1+5) = 7
7 !€ U(K^2)
In ambele cazuri, ultima cifra nu corespunde.
Concluzie : N != K^2
! inseamna 'nu'
