Arithmetical Wonderland by Unknown

Author:Unknown
Language: eng
Format: epub

Applying the second step of the B´ezoutian Algorithm, we have

1 = 5 − 2(2)

= 5(5) − 2(12)

= 5(17) − 7(12)

= 12(17) − 7(29)

= 12(75) − 31(29)

= 74(75) − 31(179)

= 74(254) − 105(179) .

A particular solution= −to 1 = 254 x = − 179 y is x = 74 and y = 105. It follows that = a particular − solution = to 4 − 254 x 179 y is x 296 and y 420. The general solution is x 296 ≤ 179 k and y 420 254 k for any integer k . Since x is non-negative, we must have 179 k 296,

4.3 Linear Diophantine Problems 103

yielding = k ≤ 1. Since= we − want = the non-negative = number x to = be as small− as = possible, we take k 1sothat x 296 179 117. For k 1, we have y 420 254 166. Thus if the

Duchess’s Cook filled the larger vessel 117 times and emptied the smaller one 166 times, there will be 4 liters of water left in the larger vessel.

Second Solution Suppose the Duchess’s Cook always poured from the smaller vessel to the larger one. Let− her fill= the smaller vessel x times and empty the larger vessel y times. Then we =− 420have 179 x+ 254 y =− 2964. From + the First Solution, the general solution to this equation ≥ is x ≤ 254 k and y 179 k . Since x is non-negative, we must have 254 k 420,

yielding = k 2. Since=− 420 we + want the= non-negative = number x to =− 296be as small + as =possible, we take k 2 so that x 508 88. For k 2, we have y 358 62. Thus if the

Duchess’s Cook filled the smaller vessel 88 times and emptied the larger one 62 times, there will be 4 liters of water left in the smaller vessel.

Example 4.3.2. The March Hare found a long row of evenly-spaced toadstools. He crouched on one of them, and started hopping from toadstool to toadstool. He found that his right leg was stronger than his left. So in one hop, he either landed on the ninth toadstool to his left, or the seventh toadstool to his right. How could he land on the toadstool immediately to the right of the one from which he started?

Solution Let x be the number of right-hops and y be the number of left-hops the March Hare had made. Then the total distance he had −travelled = to the right was 7 x and the total distance he = had travelled = to the left was 9 y . Hence 7 x 9 y 1. By inspection, a particular solution is x 4 and y 3. Hence if the March Hare hopped four times to the right and then three times to the left, he would land on the desired toadstool.

Example 4.3.3. The Mad Hatter bought several horses at $344 each and several cows at $265 each. The total amount the Mad Hatter spent on the horses was $33 more than the total amount he spent on the cows.

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