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    Hello

    Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great

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    was wrong sec
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    (Original post by Daniel Atieh)
    Hello

    Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great

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    Alright, I'm here!
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    (Original post by raniafern)
    Alright, I'm here!
    perfect
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    (Original post by Daniel Atieh)
    Hello

    Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great
    I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then. :getmecoat:

    For a pass code a 8 c d e f I got as far as:

    (10a+8)+(10c+d)+(10e+f)=80
    (100a+80+c)+(100d+10e+f)=800
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    (Original post by RDKGames)
    I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then. :getmecoat:

    For a pass code a 8 c d e f I got as far as:

    (10a+8)+(10c+d)+(10e+f)=80
    (100a+80+c)+(100d+10e+f)=800
    Haha, yup. And then, you solve the equation by eliminating 10e + f, then you eliminate c and then put them together... You get 10A + 11D + C = 72, if I'm not mistaken... I've got to solve the rest. but Ethan Hunt is on TV at the moment :love:
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    well by forming some equations and finding some restrictions you can at least shorten the trial and error procedure but I dunno if there's enough info to avoid that completely

    For instance I am able to obtain
    _ 8 _ _ 1 _
    along with some equations
    to deduce that either the first digit is 2 or it is 3, and checking out 2 that turns out to be correct giving

    283517

    I labelled it like so

    a 8 b c d e

    First thing to note is that 8 + c + e has to end in a 0. The only way this is possible is if c + e = 12.
    Similarly b + e = 10
    These lead to c - b = 2.

    We have a8b + cde = 800, and since b + e = 10, we will carry the 1 and see that 8 + d + 1 = 10, giving d = 1.

    We then note that a + c + 1 = 8, so a + c = 7.

    Similarly we have a8 + bc + 1e = 80, and carrying the 2 from 8 + c + e = 20 we get a + b + 1 + 2 = 8, so a + b = 5

    a cannot equal 1 because there is already a 1, and a cannot equal 4 because then b = 1, and there is already a 1. This leaves a =0,2,3 to check. a = 0 clearly doesn't work because then b = 5, and so e = 5 (as b + e = 10). a = 3 gives b = 2, which in turn gives c = 4, but then e = 8, which has already been used. So a = 2, from which everything else follows.

    edit: oops, I forgot you can have 2,0 back at the start there. Still, the answer is right. (and 2,0 immediately falls apart, if e = 0 then b = 10, which can't be allowed, if e = 2 then b = 8 which has been used)

    (Original post by Daniel Atieh)
    Hello Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great Name:  bmattt.PNG
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    (Original post by Daniel Atieh)
    Hello

    Any help on nailing this question using a quick approach pls? I tried to use the guessing method using different numbers, but that isn't how it is expected. So any hints or solutions would be great

    Name:  bmattt.PNG
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    You need to use some logic/deduction. Algebra is probably not the best approach.

    A 8 + B C + D E = 80

    A 8 B + C D E = 800

    Start with the first line:

    80 ends with '0' so you need 8 + C + E to end with a '0'. So C + E must end with a '2'.

    So C, E could both be 1 but all the digits are different so that's not right.
    Then C, E must add up to 12 since two single digit numbers can't add to 22.

    So you could have (C,E) = (5,7), (7,5), (6,6), (8,4), (4,8), (9,3), (3,9).

    Three of these are not possible because of duplicates.


    Try carrying on from here, If you get stuck, post all your working / thoughts.

    My answer is '7'.
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    (Original post by RDKGames)
    I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then. :getmecoat:

    For a pass code a 8 c d e f I got as far as:

    (10a+8)+(10c+d)+(10e+f)=80
    (100a+80+c)+(100d+10e+f)=800
    I'm not a fan of using these finicky 100s and stuff
    best to just write it like the by hand addition we did in primary lol, intuitive and simpler
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    (Original post by 13 1 20 8 42)
    well by forming some equations and finding some restrictions you can at least shorten the trial and error procedure but I dunno if there's enough info to avoid that completely
    (Original post by notnek)
    You need to use some logic/deduction. Algebra is probably not the best approach.
    (Original post by raniafern)
    Haha, yup. And then, you solve the equation by eliminating 10e + f, then you eliminate c and then put them together... You get 10A + 11D + C = 72, if I'm not mistaken... I've got to solve the rest. but Ethan Hunt is on TV at the moment :love:
    (Original post by RDKGames)
    I had a good think about this but I couldn't arrive to a proper method of solving this, but at first I posted that it's either 0 or 2, then I realised I'm wrong. The only mathematical thing I could think of using would be matrices but even that I'm unsure of how to use for this. 2 or 3 digits would be straight forward but 6 is something new for me. So I'll just sit tight for a proper solution until then. :getmecoat:
    Thanks a bunch for taking part, and I saw interesting replies and approaches.
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    I solved it in less than a minute through quick trial and error.

    You know that the ending digits of the latter two digit numbers have to equal 2 or 12 with means you only have to work with (0,2)(9,3)(7,5).

    You can see instantly that you can't use 0 because it being the 4th digit would require the first to be 8 and it being the 6th would require the 3rd to also be 0.

    You then know that digit 4 can't be 9 as that would make the second condition impossible and it takes seconds to test the reverse case and see that it wouldn't work as you would have to use 1 twice.

    So now you simply pick one arrangement of 7 and 5 to give you the final answer.

    I'm sure there is a far more elegant way of doing it, but this took no time at all really :dontknow:
 
 
 
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