The topic is as follows: There are three six digits, which are AbcDef, CDEfab, EFABCD. A, B, C, D, E, F represent one digit, may be any one between 1 to 9, but they are different. The three six digits are known to meet the following conditions: abcdef * 2 = cdefabcdefab * 2 = EFABCD ask a =? , B =? , C =? D =? E =? , F =? There are more people who have given the answer. The following is listed in their way: AMOU is discussed with me again, so I will talk to him. I looked at his practice, basically and I was similar to my approach. Basically, it is not allowed to take this from a 4-fold from A. But even this, basically guess, because although there is this breakthrough point, there is still a lot of the remaining possibilities, and one test is required. When I did this topic, I took a lot of time. I spent 15 minutes, so the time was not enough to have a little chaotic, so I guess it was not very order, so I didn't guess. Let's take a better way, I always feel that this practice is too high, because the first large body has a total of 18 small questions, a total of 60 minutes, so use 15 minutes to do this, even if it is It is also so harsh than this exam. SMQ provides a better way of writing, reasoning is slightly higher, guessing less, once again, let everyone depressed, and also hope more people provide better ways. AbcDef * 2 = cdefab ===> B and D are even CDEFAB * 2 = efabcd ===> a <5 c <5 ===> a = 1 a = 1 a = 2 c = 2 c = 3 c = 4 can only have these three results a = 2 c = 4 by cdefab * 2 = efabcd ===> b = 2 * D is unlikely, A must be equal to 1, by abcdef * 2 = cdefab ===> E = 5 (Because A = 1 2 * e must = 10) in CDEFAB * 2 = EFABCD ===> C = 2 and can only (b = 8 d = 6) and (b = 4 D = 8), the first group C = 3 is launched by CDEFAB * 2 = EFABCD, so b = 4 d = 8, the remaining F can be launched soon is 7 142857 YUEST also provides a detailed description of his practice, which is also very exciting .
YUEST should be a comparison with a re-mathematics, better than me, huh, huh, huh, = 0.142857142857 ... 2/7 = 0.285714285714 ... it is not difficult to 7 except 1 cycle The place is 142857 142857 * 2 = 285714 is 7 except 2 cycles Next is 3: 428571 4: 571428 5: 714285 6: 857142 But this is what I think after I am looking at Ren, I think I think the correct reasoning It should be said that it is directly that B, D is even, may be 2, 4, 6, 8 AB * 2 impossible to carry, otherwise Abcdef * 4 should be equal to 7 digits A * 4 does not carry, A may only be 1 or 2 assumptions A = 2, b is not possible to take 2, 6, 8, only 4 AB = 24 => CD = 48 or CD = 49 BC is 4 and the topic contradiction, so A = 1 EF * 2 is still More than a dozen, F necessarily carrying a to be an odd number 1, f> = 6, E may be 0 or 5, there is no 0 in the topic, so E = 5 E = 5, EF * 2 will carry, so F is odd , 7 or 9, this B = 4, D = 8 or B = 8, D = 4 has b * 2 place equal to D, so f = 7, b = 4, D = 8 ABDEF knows easy to launch C = 2AMou's girlfriend Windy provides a way, thinking is quite clear, and there is a big possibility that the similar problem is resolved again. I do this to make AB = X, CD = Y, EF = Z; four possibilities may have four probably 2z = x 2z = x 2z = x 100 2Z = x 1002y = z 2y = z according to the first form 100 2Y = Z-1 2Y = Z 992X = Y 2X = Y-1 2X = Y 2X = Y-1
According to the second form, there are four possibilities 2X = Y 2x = Y 2x = Y 100 2X = Y 1002Z = x 2z = x 100 2z = x-1 2z = x 992y = z zy = z-1 2Y = z 2Y = Z-1 Compare these combinations The combination of two equations is 2X = Y 2x = Y2Z = x 2z = x 1002y = z zy = z-1 in these two groups in the first group of inappropriate The second set of ternary first equations X = 14Y = 28Z = 57Nethermit also brought us a way: seeing your problem, let me really excite it. Move the pen, about 5 minutes, you have an answer. The solution is quite a large road, mainly by observing the relationship between the numbers, i.e. B, D is an even number, AB <25, CD <50, EF * 2 = AB or 1AB, AB × 2 = CD. COOFUCOO, I am also 81 years Have a chance to make friends. Garychan's method seems to be more regular: all see AB, CD, EF as two digits, then 2 (AB * 1000 CD * 100 EF) = CD * 10000 EF * 100 AB 2 (CD * 10000 EF * 100 ab) = EF * 10000 AB * 100 CD three variables two formulas can only be obtained. Thus, two simplicity of EF is easier to go to EF, then CD = 2ab. Get into the first concurrent and simplification, resulting in 57 * AB = 14 * EF. Obviously, AB and EF necessarily are 14 and 57 respectively. The CD is 28. The ideal state is more fortunate: I will explain, I am so solved, set AB = X, CDEF = Y, there is 2 * (10000x y) = 100Y X (first condition), so about Since its simple: 19999X = 98Y, about the simple brigade, to obtain: 2857x = 14y, so, X, Y is an integer to obtain x = 14, y = 2857, and the original number is: 142857. I have just got it out, I don't know what it is, because it is too clever, and the second condition is useless! Calculate the cat takes a dead meal! New method provided by xinfeng: The last number is F, B, D 2 * f = b 2 * b = D has FBD 1 2 4 2 4 8 3 6 2 4 8 6 5 6 2 4 7 4 8 8 6 2 9 8 6 Take a look 2 * cdefab = efabcd so 2D = f or 2D 1 = f remaining is f = 7 b = 4 D = 8 What is the way to calculate Citizen2YY: I also Putting a way. Everyone considers it. In fact, it is very simple. The original title is as follows, we have added a line at the bottom. It can guess the results.
