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By Albert A.A.

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59. (i) Find all solutions of 34x − 62y = 8 with x, y ≥ 0. (ii) Find all solutions of 62y − 34x = 8 with x, y ≥ 0. 60. Find all solutions: (i) 242x + 1870y = 66, (ii) 327x + 870y = 66 (iii) 327x + 870y = 56. 61. Find d = (3731, 1894) and write d = 3731r + 1894s where (i) r > 0 and s < 0; (ii) r < 0 and s > 0. 62. Decide if each of the following has a solution or not. If so, find the solution with the smallest possible x ≥ 0: (i) 133x + 203y = 38, (ii) 133x + 203y = 40, (iii) 133x + 203y = 42, (iv) 133x + 203y = 44.

You might guess that every fifth Fibonacci number thereafter gains another digit, and that is the case. We leave the verification as an exercise, below. Now if F5d+2 has at least d + 1 decimal digits, then any number a with d digits satisfies a < F5d+2 . So from Lam´e’s theorem we get: Corollary 15. If a < b and a has d digits, then N(a, b) ≤ (5d + 2) − 3 < 5d. The corollary shows how efficient Euclid’s Algorithm is. Even on the worst possible examples, Euclid’s Algorithm takes less than 5d steps, where d is the number of decimal digits of the smaller of the two numbers being computed.

Based on the roles of a, b, q and r in long division, we call a the divisor, b the dividend, q the quotient, and r the remainder. Proof. We prove the Division Theorem by well-ordering. Let S = {b − ax|x is a nonnegative integer and b − ax ≥ 0}. Then S is a set of nonnegative integers and is non-empty because b = b − a · 0 is in S . So by well-ordering, S has a least element r. Clearly r = b − aq for some integer q ≥ 0. We must show that 0 ≤ r < a. Since r is in S , r ≥ 0. Is r < a? If not, then r − a ≥ 0, and r − a = b − qa − a = b − (q + 1)a ≥ 0.