HCF of 210 and 55 is expressible in the form of 210 Into X + 55 into Y then find the value of x + Y
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210 = 55 * 3 + 45
55 = 45 * 1 + 10
45 = 10 * 4 + 5
10 = 5 * 2
hence 5 is the HCF.
5 = 45 - 10 * 4 substitute the values of 10 and 45 from above equations.
= 210 - 55 * 3 - (55 - 45 ) 4
= 210 - 55 * 3 - 55 * 4 + 45 * 4
= 210 - 55 * 7 + (210 - 55 * 3) * 4
= 210 * 5 - 55 * 19
so x = 5 and y = -19
and x +y = 5 -19 = -14
55 = 45 * 1 + 10
45 = 10 * 4 + 5
10 = 5 * 2
hence 5 is the HCF.
5 = 45 - 10 * 4 substitute the values of 10 and 45 from above equations.
= 210 - 55 * 3 - (55 - 45 ) 4
= 210 - 55 * 3 - 55 * 4 + 45 * 4
= 210 - 55 * 7 + (210 - 55 * 3) * 4
= 210 * 5 - 55 * 19
so x = 5 and y = -19
and x +y = 5 -19 = -14
deepthi2712:
how did q get 55:*3 again
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