Guest Posted June 16, 2009 Report Share Posted June 16, 2009 Solve this alphametic equation, where each capital letter in bold represents a different base ten digit from 1 to 9. (EEUUI)2 = RESULTING Note: While a solution is trivial with the aid of a computer program, show how to derive it without one. Quote Link to comment Share on other sites More sharing options...
0 Pickett Posted June 16, 2009 Report Share Posted June 16, 2009 Well, because it's a 5-digit number squared resulting in a 9 digit number... we know that the 5-digit number can only be between 10000 and 31622... So, that means that E must be either a 1 or 2. We also know that I cannot be 1 (because then G would be 1), and I cannot be 5 or 6 for the same reason... So, we know if I = 2, G = 4 if I = 3, G = 9 if I = 4, G = 6 if I = 7, G = 9 if I = 8, G = 4 if I = 9, G = 1 We also know that if E = 1, then I cannot be 9 (because G would have to be 1) This greatly limits the number of possible 5-digit number we can have that match that pattern (about 60): 11223, 11227, 11228 11332, 11334, 11337, 11338 11443, 11447 11552, 11553, 11554, 11557, 11558 11662, 11663, 11667, 11668 11772, 11773, 11774, 11778 11882, 11883, 11887 11992, 11993, 11994, 11997, 11998 22113, 22114, 22117, 22118 22334, 22337, 22338, 22339 22443, 22447, 22449 22553, 22554, 22557, 22558, 22559 22663, 22667, 22668, 22669 22773, 22774, 22778, 22779 22883, 22884, 22887, 22889 22993, 22994, 22997, 22998 Which, you COULD just trial and error with those results, and it wouldn't take long to find that the only answer is: (22887)2 = 523814769...I'll leave it to someone else to narrow it down more if they like... Quote Link to comment Share on other sites More sharing options...
0 Pickett Posted June 16, 2009 Report Share Posted June 16, 2009 Well, because it's a 5-digit number squared resulting in a 9 digit number... we know that the 5-digit number can only be between 10000 and 31622... So, that means that E must be either a 1 or 2. We also know that I cannot be 1 (because then G would be 1), and I cannot be 5 or 6 for the same reason... So, we know if I = 2, G = 4 if I = 3, G = 9 if I = 4, G = 6 if I = 7, G = 9 if I = 8, G = 4 if I = 9, G = 1 We also know that if E = 1, then I cannot be 9 (because G would have to be 1) This greatly limits the number of possible 5-digit number we can have that match that pattern (about 60): 11223, 11227, 11228 11332, 11334, 11337, 11338 11443, 11447 11552, 11553, 11554, 11557, 11558 11662, 11663, 11667, 11668 11772, 11773, 11774, 11778 11882, 11883, 11887 11992, 11993, 11994, 11997, 11998 22113, 22114, 22117, 22118 22334, 22337, 22338, 22339 22443, 22447, 22449 22553, 22554, 22557, 22558, 22559 22663, 22667, 22668, 22669 22773, 22774, 22778, 22779 22883, 22884, 22887, 22889 22993, 22994, 22997, 22998 Which, you COULD just trial and error with those results, and it wouldn't take long to find that the only answer is: (22887)2 = 523814769...I'll leave it to someone else to narrow it down more if they like... Going by the results I have listed above, then if E=1, then that would make R=1 (because 11998^2 = 143952004 and 11223^2 = 125955729...) So we then know that E MUST be 2...so we are down to about 30 possible numbers: 22113, 22114, 22117, 22118 22334, 22337, 22338, 22339 22443, 22447, 22449 22553, 22554, 22557, 22558, 22559 22663, 22667, 22668, 22669 22773, 22774, 22778, 22779 22883, 22884, 22887, 22889 22993, 22994, 22997, 22998 We can then determine that R MUST be 4 or 5 because 22113 * 22113 = 488984769 and 22998 * 22998 = 528908004 But it can't be 4, because since E = 2, we know that 488984769 > 42???????...so R=5 Now, it should be noted that we now know that the result begins with 52...and the UPPER bound on our results is 528908004...so we know that the number that works must be close to the upper bound (specifically within 22883 and 22998)...and really, we can get rid of 22998, since the result has zeroes in it, but we'll ignore that for now...so we are left with only these 8 possible numbers: 22883, 22884, 22887, 22889 22993, 22994, 22997, 22998 So we know that U can only be 8 or 9, and I can only be 3, 4, 7, 8 or 9... Now, let's say that G=4...then we know that I=2 or I=8...but I can't equal 2 since we know that E=2...so I=8: (22UU8)^2 = 52SULT8N4 which makes U=9: (22998)^2 = 52S9LT8N4..but we rule this one out right away, because 22998 was our upper bound of 528908004, which obviously doesn't work...so G cannot be 4, and in turn, I cannot be 8...and we also find that U cannot be 9 Since U cannot be 9, we know that U must be 8: (2288I)^2 = 52S8LTING So, now we have 4 possible answers: 22883, 22884, 22887, and 22889... We can then find out that S MUST equal 3 (since for all 4 of them, they have the same value for S)...Which means that I cannot be 3 anymore:: (2288I)^2 = 5238LTING So we only have 22884, 22887, and 22889 as our possible numbers now...We can really just start eliminating them. First assume I=4, then you have 22884*22884 which is 523677456...but we know that it must be 5238?????...so that doesn't work... Now try I=9, and you get 523906321...which again does not match 5238?????...so that only leaves I=7: (22887)^2 = 523814769 (EEUUI)^2 = RESULTING And as you can see, the result works out to be: R=5 E=2 S=3 U=8 L=1 T=4 I=7 N=6 G=9 Whew... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 17, 2009 Report Share Posted June 17, 2009 (edited) Going by the results I have listed above, then if E=1, then that would make R=1 (because 11998^2 = 143952004 and 11223^2 = 125955729...) So we then know that E MUST be 2...so we are down to about 30 possible numbers: 22113, 22114, 22117, 22118 22334, 22337, 22338, 22339 22443, 22447, 22449 22553, 22554, 22557, 22558, 22559 22663, 22667, 22668, 22669 22773, 22774, 22778, 22779 22883, 22884, 22887, 22889 22993, 22994, 22997, 22998 We can then determine that R MUST be 4 or 5 because 22113 * 22113 = 488984769 and 22998 * 22998 = 528908004 But it can't be 4, because since E = 2, we know that 488984769 > 42???????...so R=5 Now, it should be noted that we now know that the result begins with 52...and the UPPER bound on our results is 528908004...so we know that the number that works must be close to the upper bound (specifically within 22883 and 22998)...and really, we can get rid of 22998, since the result has zeroes in it, but we'll ignore that for now...so we are left with only these 8 possible numbers: 22883, 22884, 22887, 22889 22993, 22994, 22997, 22998 So we know that U can only be 8 or 9, and I can only be 3, 4, 7, 8 or 9... Now, let's say that G=4...then we know that I=2 or I=8...but I can't equal 2 since we know that E=2...so I=8: (22UU8)^2 = 52SULT8N4 which makes U=9: (22998)^2 = 52S9LT8N4..but we rule this one out right away, because 22998 was our upper bound of 528908004, which obviously doesn't work...so G cannot be 4, and in turn, I cannot be 8...and we also find that U cannot be 9 Since U cannot be 9, we know that U must be 8: (2288I)^2 = 52S8LTING So, now we have 4 possible answers: 22883, 22884, 22887, and 22889... We can then find out that S MUST equal 3 (since for all 4 of them, they have the same value for S)...Which means that I cannot be 3 anymore:: (2288I)^2 = 5238LTING So we only have 22884, 22887, and 22889 as our possible numbers now...We can really just start eliminating them. First assume I=4, then you have 22884*22884 which is 523677456...but we know that it must be 5238?????...so that doesn't work... Now try I=9, and you get 523906321...which again does not match 5238?????...so that only leaves I=7: (22887)^2 = 523814769 (EEUUI)^2 = RESULTING And as you can see, the result works out to be: R=5 E=2 S=3 U=8 L=1 T=4 I=7 N=6 G=9 Whew... Well done, and congratulations on nailing two puzzles in a row !!!! I will not post an official solution, since that would be a duplication of your method in most part barring an aspect, which I furnish hereunder in terms of comment. Each of the letters of RESULTING are different. Since each capital letter in bold represents a different base ten digit from 1 to 9, it follows that its sum of digits and therefore the number RESULTING must be divisible by 9. It then trivially follows that EEUUI is divisible by 3. This aspect will greatly reduce the total number of possibities at various stages. While on the subject, I would like to invite your attention to this puzzle, which you may find interesting. Edited June 17, 2009 by K Sengupta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 17, 2009 Report Share Posted June 17, 2009 (edited) Since EEUUI is 5 digit and RESULTING is 9 digit, E is either 1 or 2 If E is 1, R must be 1.. so E must be 2 Since E is 2, R must be 4 or 5 However, with EE starting, R can only be 5 as RE must be either 42 or 52 So, RE is 52.. which means U is either 9 or 8 Now, I can take values 3,4,7,8,9 only and corresponding values of G are 9,6,9,4,1 Now, RESULTING is divisible by 9 (all digits 1 to 9 used)... so EEUUI must be divisible by 3. Then E+E+U+U+I must be divisible by 3 If E is 9, then E+E+U+U = 2+2+9+9 = 22... For EEUUI to be divisible by 3, I must be 8 If E is 8, then E+E+U+U = 2+2+8+8 = 20... For EEUUI to be divisible by 3, I must be 4 or 7 so the possible numbers are 22998, 22884 and 22887 Of these only 22887 satisfies the conditions. Hopefully, someone can find a better way to eliminate the last 2 choices also in a more elegant way (without having to do calculations) EDIT: Oops, I didnt see K Sengupta had already posted the funda behind this. Edited June 17, 2009 by DeeGee Quote Link to comment Share on other sites More sharing options...
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Solve this alphametic equation, where each capital letter in bold represents a different base ten digit from 1 to 9.
(EEUUI)2 = RESULTING
Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.
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