560 n E10 mod 70 n 3 mod 80 and n E 2 mod 90 SolutionSolutio

560

n E10 mod 70, n 3 mod 80, and n E 2 mod 90

Solution

n = 1mod7 ……… (1)

n = 3 mod 8 ……… (2) and

n = 2 mod 9 ……… (3)

(1) => n = 7k1 + 1 ….(4)

(2) => n = 8k2 + 3 ….(5)

(4) and (5) => 7k1 + 1 = 8k2 + 3 or 8k2 – 7k1 = - 3 + 1 = - 2 ……. (6)

Dividing (6) by 7 (smaller of the coefficients) and equating the remainders,

R(k2/7) = R(- 2/7) = 5 [note R(- 2/7) = 5 since 2 = 7- 5] => k2 = 5, 12, 19. 26, 33, 40, …….. (7)

(3) => n = 9k3 + 2 ……. (8)

(5) and (8) => 8k2 + 3 = 9k3 + 2 or 9k3 – 8k2 = 1…… (8)

Dividing (8) by 9 (so that k3 is eliminated and k2 remains) and equating the remainders,

- R(8k2/9) = R(1/9) or – R{(9k2 – k2)/9} = 1or – R{(– k2)/9} = 1 or R( k2)/9 = 1 =>

K2 = 1, 10, 19, 28, 37. 46, …… (9)

Visual comparison of (7) and (9) trivially shows that

k2 = 19 => n = (8 x 19) + 3 = 155 ANSWER.