Question

OV - (a) Suppose that the GCIS of two numbers is 10 and their product is 4000 find the Lcm of the two numbers. solve the following system of congruenice x=4 (mods) x= 2 (mod3)

Answer 1

Answer: @ we have to find two number whose product 4000 tposand OCD is 10 and then find Lom of these two let az 10 and b=400 Ca, b) = Clo, 400) = 10 and a. b= 10x400 = 4.000 Lom = axb Cabi 1. [a,b] = [10.40] = To 4000 = 400 = [ 10,400] = 400 so we solve the following congruencess by choinnese chinese remainder theorem Given x=4 cmods) x = 2 (mod 3) bet ai=4 and a2=2 and mi=5 and m₂=3 All min are relatively prime s) i 5 f3 are relatively prime lie.(5,3 Memixm2=5x3 =15 Again mi= m = 15 -3 & M = m = 15 = 5

consider (mi, mi) = 3,5) "we find G CD & 3 85 by division algorithm. : 5=3*1+2 consider (M2, M2) = (53) 3= 2x1+1 :523 x 1+2 2 = 1x2 to 3 =2x 1+1 - (8 3,5) = 1 2 = 1X2 to . - 1 = 3-2x 1 = 3-(5-3x10x1 :. 1= 3-2x1 2. = 3-(5-3x10x1 = 3x2 - 581 = 3 x 2 - 5X I = -5X1+ 3 X2 z - 5x t +3 X 2 1 = 5X(-1) + 3 X (2) 1 = 5*(-D + 3 X (2) b =2 No= mi .a; bj = mia, bit Mr A2 62 = 3 X 4 X 2 +5X2 X (-1) = 24-10=14 xo=14 Ko= 14.(modus) - which is required solution
By using the definition of LCM and Chinese remainder theorem.i was solved this question.