Commander Posted December 10, 2014 Posted December 10, 2014 Hi All,Here is an interesting Puzzle ! ADCDE x 4 = EDCBA where A B C D E stands for a decimal symbol each[one of 0,1,2,3,4,5,6,7,8,9]Thus ADCDE and EDCBA are each a 5 digit decimal numberCan you find the answer ? ABCDE x 4 ............. EDCBA .............What is A, B, C, D, E ?................................................................................................For a Set of INTERESTING PUZZLES Click on url deleted And Click on ' Interesting Puzzles ' under ' Family Corner ' Good Luck for Solving them !!
fiveworlds Posted December 10, 2014 Posted December 10, 2014 (edited) I bet one of the integers in both is zero. ADCDE x 4 = EDCBA AD0DE x 4 = ED0BA Edited December 10, 2014 by fiveworlds 1
studiot Posted December 11, 2014 Posted December 11, 2014 (edited) Fiveworlds, since both have 5 digits A cannot be greater than 2 ie A = 0 or A = 1 or A = 2A=0 leads to the trivial solution (00000) x 4 = (00000), there are no other solutions containing zero. (21978) x 4 = 87912 Edited December 11, 2014 by studiot 1
swansont Posted December 11, 2014 Posted December 11, 2014 ! Moderator Note Link removed. Please read the rules and note 2.7, which states (in part) Advertising and spam is prohibited. We don't mind if you put a link to your noncommercial site (e.g. a blog) in your signature and/or profile, but don't go around making threads to advertise it. ADCDE x 4 = EDCBA … ABCDE x 4 ............. EDCBA ............. Which is it supposed to be, ABCDE or ADCDE? 1
Robittybob1 Posted December 11, 2014 Posted December 11, 2014 (edited) ! Moderator Note Link removed. Please read the rules and note 2.7, which states (in part) Advertising and spam is prohibited. We don't mind if you put a link to your noncommercial site (e.g. a blog) in your signature and/or profile, but don't go around making threads to advertise it. Which is it supposed to be, ABCDE or ADCDE? Starting from a five digit number and then multiplied by 4 still equals a 5 digit answer that first digit can either be 1 or 2 but not 3 Let's assume it is ABCDE and all digits different. How many different numbers can there be under 25000? If we allow A to be 0 then ABCDE is really only a 4 digit number which would imply E has to be 5, but any 5 digit number starting with 5 divided by 4 will be over 10,000 so both number are truly 5 digits i.e. A <>0. Edited December 11, 2014 by Robittybob1
Commander Posted December 11, 2014 Author Posted December 11, 2014 (edited) Hi Friends, This is MY OWN PUZZLE [means created by me] as I wanted to create it in the form ABCDE multiplied by X = EDCBA ! Therefore I tried to solve it myself and found the Solution for ABCDE x 4 = EDCBA I have not yet analyzed and found there is any solution for ABCDE x n = EDCBA where n is 2,3, 5 or 6 !! Also, I did find one Solution but did not search for any more Solutions. I need to now look for the notes where I wrote that Solution or Solve it once again myself !!! >>>>>>>>>>>>>>>>>>>>> THERE IS A SOLUTION 100 % and don't lose heart !!!! As a CLUE I can add that THERE IS NO LEADING ZERO in any of the 5 Digit numbers in the Equation. ie. Neither A or E is 0. >>>>>>>>>>>>>>>>>>>> PS : The URL quoted in my post is nothing but my own Personal Website and the same can be accessed from my Profile ! Edited December 11, 2014 by Commander
Robittybob1 Posted December 11, 2014 Posted December 11, 2014 Hi Friends, This is MY OWN PUZZLE [means created by me] as I wanted to create it in the form ABCDE multiplied by X = EDCBA ! Therefore I tried to solve it myself and found the Solution for ABCDE x 4 = EDCBA I have not yet analyzed and found there is any solution for ABCDE x n = EDCBA where n is 2,3, 5 or 6 !! Also, I did find one Solution but did not search for any more Solutions. I need to now look for the notes where I wrote that Solution or Solve it once again myself !!! >>>>>>>>>>>>>>>>>>>>> THERE IS A SOLUTION 100 % and don't lose heart !!!! Are they all different values or can they be the same number i.e. can both A and B = 2?
Commander Posted December 11, 2014 Author Posted December 11, 2014 (edited) Are they all different values or can they be the same number i.e. can both A and B = 2? YES EACH ALPHABET REPRESENTS A UNIQUE/DIFFERENT DIGIT ! None of A,B,C,D,E are equal to one another !! ! Moderator Note Link removed. Please read the rules and note 2.7, which states (in part) Advertising and spam is prohibited. We don't mind if you put a link to your noncommercial site (e.g. a blog) in your signature and/or profile, but don't go around making threads to advertise it. Which is it supposed to be, ABCDE or ADCDE? ABCDE Edited December 11, 2014 by Commander 1
studiot Posted December 11, 2014 Posted December 11, 2014 (edited) Thank you swansont for sorting my spoiler. Robbitybob, It is not necessary to know if all the digits are unique in order to solve this one. You can work on this one from the outside in, like peeling an onion, First A, then E, then B then............................. Thank you commander for a neat problem, I assume you have established a set of simultaneous equations as a method since you mention the general case. +1 will head you off into the right direction (the green) I look forward to more useful and entertaining posts in the future. Edited December 11, 2014 by studiot
Commander Posted December 11, 2014 Author Posted December 11, 2014 Thank you swansont for sorting my spoiler. Robbitybob, It is not necessary to know if all the digits are unique in order to solve this one. You can work on this one from the outside in, like peeling an onion, First A, then E, then B then............................. Thank you commander for a neat problem, I assume you have established a set of simultaneous equations as a method since you mention the general case. +1 will head you off into the right direction (the green) I look forward to more useful posts and entertaining in the future. Hi, Thanks. To be fair to all who are trying hard to get the solution, I must hold on for a while before putting up the Solution. Yes, as I said this is a straight forward puzzle with a clear answer !
imatfaal Posted December 11, 2014 Posted December 11, 2014 Commander - Studiot has already solved it; you can click where it says Spoiler in his post here We use Spoilers so that one member can demonstrate that they have solved the teaser whilst hiding that solution from others who may wish to continue to search demo 21978*4=87912 As far as I can tell 4 is the only multiplier that fits the bill strictly. If repeated digits are allowed then multiplier 9 will work (10989) and multiplier 1 will work for all palindromic numbers trivially.
Commander Posted December 11, 2014 Author Posted December 11, 2014 Fiveworlds, since both have 5 digits A cannot be greater than 2 ie A = 0 or A = 1 or A = 2 A=0 leads to the trivial solution (00000) x 4 = (00000), there are no other solutions containing zero. (21978) x 4 = 87912 Oh yes, well done - Congratulations Studiot ! It was indeed Bright !! Commander - Studiot has already solved it; you can click where it says Spoiler in his post here We use Spoilers so that one member can demonstrate that they have solved the teaser whilst hiding that solution from others who may wish to continue to search demo 21978*4=87912 As far as I can tell 4 is the only multiplier that fits the bill strictly. If repeated digits are allowed then multiplier 9 will work (10989) and multiplier 1 will work for all palindromic numbers trivially. Hi, yes Thank you for the info. Studio has solved it quite quickly and the Solution may be unique !
studiot Posted December 11, 2014 Posted December 11, 2014 There's invariably more than one method in these things and the real fun comes in seeing how someone else has done it.
Robittybob1 Posted December 12, 2014 Posted December 12, 2014 Hi Friends, This is MY OWN PUZZLE [means created by me] as I wanted to create it in the form ABCDE multiplied by X = EDCBA ! Therefore I tried to solve it myself and found the Solution for ABCDE x 4 = EDCBA I have not yet analyzed and found there is any solution for ABCDE x n = EDCBA where n is 2,3, 5 or 6 !! Also, I did find one Solution but did not search for any more Solutions. I need to now look for the notes where I wrote that Solution or Solve it once again myself !!! >>>>>>>>>>>>>>>>>>>>> THERE IS A SOLUTION 100 % and don't lose heart !!!! As a CLUE I can add that THERE IS NO LEADING ZERO in any of the 5 Digit numbers in the Equation. ie. Neither A or E is 0. >>>>>>>>>>>>>>>>>>>> PS : The URL quoted in my post is nothing but my own Personal Website and the same can be accessed from my Profile ! I didn't find any other solutions either. for 2,3,5, 6,7, 8. only 1 solution for 4.
Commander Posted December 12, 2014 Author Posted December 12, 2014 (edited) I didn't find any other solutions either. for 2,3,5, 6,7, 8. only 1 solution for 4. Yes, thank you. I once again solved it trying all possibilities and found that the Solution found by Studiot is the only Solution and unique. I also agree with you that the multiplier needs to be only 4 and nothing else PS : In fact I have not yet verified it for 2 or 3 in place of 4 But 5,6,7 or 8 might be eliminated as if the multiplier is any of them A needs to be only 1 which straight away eliminates 5 , 6 and 8 but 7 needs to be investigated. Edited December 12, 2014 by Commander
Robittybob1 Posted December 12, 2014 Posted December 12, 2014 Yes, thank you. I once again solved it trying all possibilities and found that the Solution found by Studiot is the only Solution and unique. I also agree with you that the multiplier needs to be only 4 and nothing else So how did you work it out? I used a macros that that looked at randon numbers and checked them for validity.
studiot Posted December 12, 2014 Posted December 12, 2014 (edited) Here is the next stage of my reasoning; I have already indicated that A must be 1 or 2. Now consider multiplying E by 4. E is the only digit that does not entail a potential carry digit, so the result of the multiplication of the whole number by 4 has the same last digit as 4E. The result of the multiplication must end in either 1 or 2, since the last digit of 4E = A But there is no number such that the last digit of 4E can be 1 so A must = 2. So 4E ends in 2. There (4x3=12) and (4x8=32) both end in 2 so E is either 3 or 8 Since A = 2; (4A + carry from 4B) can only be 8 or 9, But we are told the leading digit is E so E = 8. Since E = 8 there is no carry on multiplying B by 4, so B must be either 0, 1 or 2 Edited December 12, 2014 by studiot
Commander Posted December 12, 2014 Author Posted December 12, 2014 So how did you work it out? I used a macros that that looked at randon numbers and checked them for validity. Well I solved just arithmetically with all possibilities taken care of such as : A can not be anything but 2 and therefore E has to be only 8 as given below. A can not be 1, 3 , 5 ,7 or 9 which is trivial as no product of 4xE will end with those digits. A can not be 0 because if it is so the result can not be in 5 digits. Also as per clue no leading zero. A can not be more than 2 in which case the result will exceed 5 digits. So A is 2 ! For A being 2 E has to be either 3 or 8 with 4XE either 12 or 32 Also as 4A should not exceed 9 [if so it goes into 6 digits] and as 4A is only 8 as we decided A = 2 already E can be only 8 or with Carry from 4xC can be 9. So, E is either 8 or 9 but if E = 9 4xE can not end with 2 and therefore with logic from either end E = 8 Thereafter I enumerate the possibilities of B, D as well as C and checked that apart from 21978 x 4 = 87912 I don't see any other set fulfilling the results !! Here is the next stage of my reasoning; I have already indicated that A must be 1 or 2. Now consider multiplying E by 4. E is the only digit that does not entail a potential carry digit, so the result of the multiplication of the whole number by 4 has the same last digit as 4E. The result of the multiplication must end in either 1 or 2, since the last digit of 4E = A But there is no number such that the last digit of 4E can be 1 so A must = 2. So 4E ends in 2. There (4x3=12) and (4x8=32) both end in 2 so E is either 3 or 8 Since A = 2; (4A + carry from 4B) can only be 8 or 9, But we are told the leading digit is E so E = 8. Since E = 8 there is no carry on multiplying B by 4, so B must be either 0, 1 or 2 Yes, the correct reasoning !
studiot Posted December 12, 2014 Posted December 12, 2014 (edited) Thereafter I enumerate the possibilities of B, D as well as C and checked that apart from 21978 x 4 = 87912 I don't see any other set fulfilling the results !! You can carry my logic all the way in from the outside, as I said before, working B, then D then finally C using the carry conditions each time. I made a start of B for you in post#17 Edited December 12, 2014 by studiot 1
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