Question

if the hcf of 210 and 55 is expressible in the form 210 × 5 + 55y find y

Answer 1

HCF= 5
therefore, 210*5-55y= 5
=> y is -19
hope it helped u

Answer 2

Heya folk!!

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


Hope this helps u plzz mark it as brainliest