If the hcf of 210and 55 is express in the form 210 ×5+55y then find y
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hcf(210,55)=5
5=210*5+55y
5=1050+55y
55y=-1045
y=-1045/55=-19
5=210*5+55y
5=1050+55y
55y=-1045
y=-1045/55=-19
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