Question

express the HCF of 408 and 1032 is expressible in the form of 408x +1032y where x, y are integers in two different ways.

Answer 1

Since 1032>408
So by using Euclid's division algorithm
1032 = 408*2+216
408 = 216*1+192
216 = 192*1+24
192 = 24*8+0
So the HCF of (1032,408) is 24.
Now,
24 = 216-192*1
= 216-(408-216*1)*1
= 216*1-408*1+216*1
= 216*2- 408*1
= (1032-408*2)*2- 408*1
= 1032*2-408*4-408*1
= 1032*2 - 408*5
= 408*(-5) + 1032*
hence x= -5 & y= 2

Answer 2

Step-by-step explanation:

• since 1032>408
• so by using Euclid's division algorithm
• 1032 =408*2+216
• 408 =216*1 +192
• 216 =192*1 +24
• 192 =24*8 +0
• so the H.C.F of (1032,408) is 24
• Now,
• 24 = 216 -192*1
• =216 -(408 -216*1)*1
• =216*1 -408*1 +216*1
• =216*2 -408*1
• =(1032 -408*2)*2 -408*1
• =1032*2 -408*4 -408*1
• =1032*2 -408*5
• =408*(-5) +1032*
• Hence x=-5 y=2