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Explore a diophantine equation (Posted on 2023-08-14) |
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Given:
2xy+3x+2y=42
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Provide all solutions, allowing non-negative values only.
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What two-digit numbers can replace 42, so that there will still be non negative solutions for the bolded equation?
Solution
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| Comment 2 of 6 |
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Solution to part I: Factor the expression. 2xy + 2y + 3x = 42 2y(x+1) + 3(x+1) = 45 (x+1)(2y+3) = 45 A * B = N+3 A B (x,y) (x,y) for -A*-B 1 45 (0,21) (-2,-24) 3 15 (2,6) (-4,-9) 5 9 (4,3) (-6,-6) 9 5 (8,1) (-10,-4) 15 3 (14,0) (-16,-3) 45 1 (44,-1) (-46,-2) (0,21)(2,6)(4,3)(8,1)(14,0)
Solution to part II: In general, considering the factors to be in ordered pairs, If 2-digit N replaces 42, then (x+1)(2y+3) = N+3
N is from 10 to 99 (2-digit number) So N+3 is from 13 to 102 (the n in my code is really N+3)
N+3 [x,y] N 13 [0, 5] 10 14 [1, 2] 11 15 [0, 6] 12 15 [2, 1] 12 15 [4, 0] 12 17 [0, 7] 14 18 [1, 3] 15 18 [5, 0] 15 19 [0, 8] 16 20 [3, 1] 17 21 [0, 9] 18 21 [2, 2] 18 21 [6, 0] 18 22 [1, 4] 19 23 [0, 10] 20 24 [7, 0] 21 25 [0, 11] 22 25 [4, 1] 22 26 [1, 5] 23 27 [0, 12] 24 27 [2, 3] 24 27 [8, 0] 24 28 [3, 2] 25 29 [0, 13] 26 30 [1, 6] 27 30 [5, 1] 27 30 [9, 0] 27 31 [0, 14] 28 33 [0, 15] 30 33 [2, 4] 30 33 [10, 0] 30 34 [1, 7] 31 35 [0, 16] 32 35 [4, 2] 32 35 [6, 1] 32 36 [3, 3] 33 36 [11, 0] 33 37 [0, 17] 34 38 [1, 8] 35 39 [0, 18] 36 39 [2, 5] 36 39 [12, 0] 36 40 [7, 1] 37 41 [0, 19] 38 42 [1, 9] 39 42 [5, 2] 39 42 [13, 0] 39 43 [0, 20] 40 44 [3, 4] 41 45 [0, 21] 42 45 [2, 6] 42 45 [4, 3] 42 45 [8, 1] 42 45 [14, 0] 42 46 [1, 10] 43 47 [0, 22] 44 48 [15, 0] 45 49 [0, 23] 46 49 [6, 2] 46 50 [1, 11] 47 50 [9, 1] 47 51 [0, 24] 48 51 [2, 7] 48 51 [16, 0] 48 52 [3, 5] 49 53 [0, 25] 50 54 [1, 12] 51 54 [5, 3] 51 54 [17, 0] 51 55 [0, 26] 52 55 [4, 4] 52 55 [10, 1] 52 56 [7, 2] 53 57 [0, 27] 54 57 [2, 8] 54 57 [18, 0] 54 58 [1, 13] 55 59 [0, 28] 56 60 [3, 6] 57 60 [11, 1] 57 60 [19, 0] 57 61 [0, 29] 58 62 [1, 14] 59 63 [0, 30] 60 63 [2, 9] 60 63 [6, 3] 60 63 [8, 2] 60 63 [20, 0] 60 65 [0, 31] 62 65 [4, 5] 62 65 [12, 1] 62 66 [1, 15] 63 66 [5, 4] 63 66 [21, 0] 63 67 [0, 32] 64 68 [3, 7] 65 69 [0, 33] 66 69 [2, 10] 66 69 [22, 0] 66 70 [1, 16] 67 70 [9, 2] 67 70 [13, 1] 67 71 [0, 34] 68 72 [7, 3] 69 72 [23, 0] 69 73 [0, 35] 70 74 [1, 17] 71 75 [0, 36] 72 75 [2, 11] 72 75 [4, 6] 72 75 [14, 1] 72 75 [24, 0] 72 76 [3, 8] 73 77 [0, 37] 74 77 [6, 4] 74 77 [10, 2] 74 78 [1, 18] 75 78 [5, 5] 75 78 [25, 0] 75 79 [0, 38] 76 80 [15, 1] 77 81 [0, 39] 78 81 [2, 12] 78 81 [8, 3] 78 81 [26, 0] 78 82 [1, 19] 79 83 [0, 40] 80 84 [3, 9] 81 84 [11, 2] 81 84 [27, 0] 81 85 [0, 41] 82 85 [4, 7] 82 85 [16, 1] 82 86 [1, 20] 83 87 [0, 42] 84 87 [2, 13] 84 87 [28, 0] 84 88 [7, 4] 85 89 [0, 43] 86 90 [1, 21] 87 90 [5, 6] 87 90 [9, 3] 87 90 [17, 1] 87 90 [29, 0] 87 91 [0, 44] 88 91 [6, 5] 88 91 [12, 2] 88 92 [3, 10] 89 93 [0, 45] 90 93 [2, 14] 90 93 [30, 0] 90 94 [1, 22] 91 95 [0, 46] 92 95 [4, 8] 92 95 [18, 1] 92 96 [31, 0] 93 97 [0, 47] 94 98 [1, 23] 95 98 [13, 2] 95 99 [0, 48] 96 99 [2, 15] 96 99 [8, 4] 96 99 [10, 3] 96 99 [32, 0] 96 100 [3, 11] 97 100 [19, 1] 97 101 [0, 49] 98 102 [1, 24] 99 102 [5, 7] 99 102 [33, 0] 99
2-digit numbers that work [10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99] There are 87 of these, including 42
-------------- def factorPairs(n): ans = [] for i in range(1,n+1): if (n/i)%1 == 0: ans.append([i,int(n/i)]) return ans
nlist = [] print('N+3 [x,y] N ') for n in range(13,103): # comment out one of these lines # for n in range(45,46): # comment out one of these lines for pair in factorPairs(n): y = (pair[1] - 3)/2 if y%1 != 0: continue x = pair[0] - 1 if x<0 or y<0: continue y = int(y) print(n, [x,y], 2*x*y+3*x+2*y ) if n-3 not in nlist: nlist.append(n-3) print(nlist ) print(len(nlist))
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Posted by Larry
on 2023-08-14 10:24:03 |
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