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Find the value of y if the HCF of 210 and 55 is expressible in the form 210 x5 + 55y |
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Answer» 210 = 55 * 3 + 4555 = 45 * 1 + 1045 = 10 * 4 + 510 = 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 Let us first find the HCF of 210 and 55. Applying Euclid division lemna on 210 and 55, we get 210 = 55 × 3 + 45 ....(1) Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get 55 = 45 × 1 + 10 .... (2) Again, considering the divisor 45 and remainder 10 and applying division lemna, we get 45 = 4 × 10 + 5 .... (3) We now, consider the divisor 10 and remainder 5 and applying division lemna to get 10 = 5 × 2 + 0 .... (4) We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55. ∴ 5 = 210 × 5 + 55y ⇒ 55y = 5 - 1050 = -1045 ⇒ y = -19 |
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