Question

If the HCF of 210 and 55 is expressible in form 210×5+55y,find Y

Answer 1

Answer:

Step-by-step explanation:

we know the formula a = bq + r

210 = 55 × 3 +45 say it is eq . 1)

55 = 45 ×1 + 10 2)

45 = 4×10 +5 3)

now we consider the divisor 10 and the remainder 5 and apply DL to get

10 = 5×2 +0

we observe here that r =0 and our present divisor which is 5 will be it HcF

A/Q

5 = 210 ×5 + 55y

55y = 5 - 210×5

= 5 - 1050

y = - 1045 /55

y=-19

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