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Sir Col
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Distinct Division
« on: Aug 29th, 2007, 7:20am » |
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Given that a, b, and c are different digits, how many distinct solutions exist in the quotient: baba / a = aca? (Note that baba represents a 4-digit number and not the product a2b2)
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Grimbal
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Re: Distinct Division
« Reply #1 on: Aug 29th, 2007, 7:37am » |
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Let's write it: a·aca = baba Considering the last digit, a can only be 1, 5, 6, and it is obviously not 1. 5*5c5 = 2525 + 50·c works only with c=0 6*6c6 = 3636 + 60·c works only with c=0 So, I only see (a,b,c) = (5,2,0) and (6,3,0)
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towr
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Re: Distinct Division
« Reply #2 on: Aug 29th, 2007, 8:25am » |
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a * aca = baba baba = 101 * ba 101 * ba = a * aca for a < 101, 101 must divide aca, so then c = 0 so ba = a * a In decimal,that leaves us 3,6 and 2,5 for b,a For other bases I'll have to take a better look. and of course for bases in which a can be greater than 101, the story can also be different
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towr
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Re: Distinct Division
« Reply #3 on: Aug 29th, 2007, 8:39am » |
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hidden: | All pairs a,b in (natural) number bases smaller or equal to 101 6: (3, 1), (4, 2) 10: (5, 2), (6, 3) 12: (4, 1), (9, 6) 14: (7, 3), (8, 4) 15: (6, 2), (10, 6) 18: (9, 4), (10, 5) 20: (5, 1), (16, 12) 21: (7, 2), (15, 10) 22: (11, 5), (12, 6) 24: (9, 3), (16, 10) 26: (13, 6), (14, 7) 28: (8, 2), (21, 15) 30: (6, 1), (10, 3), (15, 7), (16, 8), (21, 14), (25, 20) 33: (12, 4), (22, 14) 34: (17, 8), (18, 9) 35: (15, 6), (21, 12) 36: (9, 2), (28, 21) 38: (19, 9), (20, 10) 39: (13, 4), (27, 18) 40: (16, 6), (25, 15) 42: (7, 1), (15, 5), (21, 10), (22, 11), (28, 18), (36, 30) 44: (12, 3), (33, 24) 45: (10, 2), (36, 28) 46: (23, 11), (24, 12) 48: (16, 5), (33, 22) 50: (25, 12), (26, 13) 51: (18, 6), (34, 22) 52: (13, 3), (40, 30) 54: (27, 13), (28, 14) 55: (11, 2), (45, 36) 56: (8, 1), (49, 42) 57: (19, 6), (39, 26) 58: (29, 14), (30, 15) 60: (16, 4), (21, 7), (25, 10), (36, 21), (40, 26), (45, 33) 62: (31, 15), (32, 16) 63: (28, 12), (36, 20) 65: (26, 10), (40, 24) 66: (12, 2), (22, 7), (33, 16), (34, 17), (45, 30), (55, 45) 68: (17, 4), (52, 39) 69: (24, 8), (46, 30) 70: (15, 3), (21, 6), (35, 17), (36, 18), (50, 35), (56, 44) 72: (9, 1), (64, 56) 74: (37, 18), (38, 19) 75: (25, 8), (51, 34) 76: (20, 5), (57, 42) 77: (22, 6), (56, 40) 78: (13, 2), (27, 9), (39, 19), (40, 20), (52, 34), (66, 55) 80: (16, 3), (65, 52) 82: (41, 20), (42, 21) 84: (21, 5), (28, 9), (36, 15), (49, 28), (57, 38), (64, 48) 85: (35, 14), (51, 30) 86: (43, 21), (44, 22) 87: (30, 10), (58, 38) 88: (33, 12), (56, 35) 90: (10, 1), (36, 14), (45, 22), (46, 23), (55, 33), (81, 72) 91: (14, 2), (78, 66) 92: (24, 6), (69, 51) 93: (31, 10), (63, 42) 94: (47, 23), (48, 24) 95: (20, 4), (76, 60) 96: (33, 11), (64, 42) 98: (49, 24), (50, 25) 99: (45, 20), (55, 30) 100: (25, 6), (76, 57) |
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« Last Edit: Aug 29th, 2007, 8:41am by towr » |
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brown_eye
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Re: Distinct Division
« Reply #4 on: Aug 31st, 2007, 6:01am » |
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reduce it to (10b)-(10/101)*c = a(a-1)... that can be an approach .....
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Grimbal
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Re: Distinct Division
« Reply #5 on: Aug 31st, 2007, 6:15am » |
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It would be 10/101*ac, but it comes to the same.
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