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Another Math Problem

2»

Comments

  • #32
    yerayeah
    Closed Accounts Posts: 804 ✭✭✭yerayeah


    CA+CB=CD

    CxD/B is odd by even so it's an even number.
    Therefore CA must be even as only two evens can give an even answer.
    Therefore A must be even as for CA to be even, A must be even as C is odd.

    C(A+B)=C(D)
    A+B=D

    Since they're all different numbers, only 2+4=6 fits the bill, so D=6 and you don't know what the rest are. My take on it anyway!


  • #33
    bowsie010
    Registered Users, Registered Users 2 Posts: 168 ✭✭bowsie010


    Your child gets some faarce hard homework.


  • #34
    fergalr
    Registered Users, Registered Users 2 Posts: 1,922 ✭✭✭fergalr


    Sometimes searching problems like this are best solved on computer.
    Computers are good at trying a lot of combinations of something fast.

    • C is greater than 1.
    • C is an odd number.
    • B and D are even Numbers.

    • CA + CB=DC
    Here are some numbers that meet those criteria, assuming CA doesn't mean C*A, but instead A concatenated onto C:
    1,0,1,2
    1,0,21,4
    1,2,3,6
    1,2,63,12
    1,4,5,10
    1,6,7,14
    1,8,9,18
    1,10,1,12
    1,12,3,34
    1,14,5,56
    1,16,7,78
    1,20,31,34
    1,22,53,58
    1,24,75,82
    1,30,1,14
    .
    .
    .
    3,0,3,6
    Single Digit:

    1,0,1,2
    1,2,3,6
    3,0,3,6
    different Single Digit

    1,2,3,6


    Numbers in the range -9 <-> 0 don't satisfy it either.
    Have to say, its actually quite robustly specified for a problem like this!


  • #35
    x in the city
    Closed Accounts Posts: 7,132 ✭✭✭x in the city


    fergalr wrote: »
    Sometimes searching problems like this are best solved on computer.
    Computers are good at trying a lot of combinations of something fast.



    Here are some numbers that meet those criteria, assuming CA doesn't mean C*A, but instead A concatenated onto C:
    1,0,1,2
    1,0,21,4
    1,2,3,6
    1,2,63,12
    1,4,5,10
    1,6,7,14
    1,8,9,18
    1,10,1,12
    1,12,3,34
    1,14,5,56
    1,16,7,78
    1,20,31,34
    1,22,53,58
    1,24,75,82
    1,30,1,14
    .
    .
    .
    3,0,3,6



    1,0,1,2
    1,2,3,6
    3,0,3,6



    1,2,3,6


    Numbers in the range -9 <-> 0 don't satisfy it either.
    Have to say, its actually quite robustly specified for a problem like this!

    surprised of the intelligence of (some) of the ah motley crew tbh...:P


  • #36
    Doc
    Registered Users, Registered Users 2 Posts: 2,670 ✭✭✭Doc


    I was always told that if you have two numbers represented by letters such as A and B and you show them in a problem like this:

    AB + B = C

    The AB means that you multiply A by B.

    So that if A=4 and B=2 the above problem would be…

    (4x2) + 2 = C => C = 10

    The way you have solved OP's question this AB would be 42.

    So...

    42 + 2 = C => C = 44

    Which I believe is incorrect.

    Am I wrong???


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  • #37
    mojesius
    Registered Users, Registered Users 2 Posts: 4,394 ✭✭✭mojesius


    Thank jeebaz I don't have to do maths anymore.


  • #38
    D-Generate
    Registered Users, Registered Users 2 Posts: 2,947 ✭✭✭D-Generate


    Wow some people went to such lengths to find out!

    C > 1 and C is odd leaving 3, 5 ,7, 9. Since we end up with a 2 digit number at the end after adding CA + CB then it can't be greater than 3.

    3A + 3B = D3

    Now to get 3 we have to add 1,2 or else we have a carry or we are using a 3 again in the case of 3,0.

    So A = 1, B = 2, C = 3 and D = 6.

    However this is supplementary to the answer as the question only asks for D so answer is D is 6.

    I am gutted that i figured it out in no time at all and was excited to be first to have the answer but alas was stripped of this accolade.


  • #39
    D-Generate
    Registered Users, Registered Users 2 Posts: 2,947 ✭✭✭D-Generate


    Doc wrote: »
    I was always told that if you have two numbers represented by letters such as A and B and you show them in a problem like this:

    AB + B = C

    The AB means that you multiply A by B.

    So that if A=4 and B=2 the above problem would be…

    (4x2) + 2 = C => C = 10

    The way you have solved OP's question this AB would be 42.

    So...

    42 + 2 = C => C = 44

    Which I believe is incorrect.

    Am I wrong???

    Well if you did it that way then wouldn't it be
    C(A + B) = C(D)
    A + B = D

    4, 2, 6
    6, 2, 8
    2, 4, 6
    2, 6, 8

    So those are the valid solutions in that instance that satisfy criteria.


  • #40
    Doc
    Registered Users, Registered Users 2 Posts: 2,670 ✭✭✭Doc


    D-Generate wrote: »
    Well if you did it that way then wouldn't it be
    C(A + B) = C(D)
    A + B = D

    4, 2, 6
    6, 2, 8
    2, 4, 6
    2, 6, 8

    So those are the valid solutions in that instance that satisfy criteria.

    That’s my point you took CA + CB=DC to mean (C is the first part of the number with A as the second) plus (C as the first part of the number with B as the second) is equal to (D as the first part of the letter with C as the second) then you can come up with a definitive answer for the value of D.

    yerayeah's way of answering is correct. You got the right answer but the wrong way


  • #41
    Adyx
    Registered Users, Registered Users 2 Posts: 2,997 ✭✭✭Adyx


    D-Generate wrote: »
    Wow some people went to such lengths to find out!

    C > 1 and C is odd leaving 3, 5 ,7, 9. Since we end up with a 2 digit number at the end after adding CA + CB then it can't be greater than 3.

    3A + 3B = D3

    Now to get 3 we have to add 1,2 or else we have a carry or we are using a 3 again in the case of 3,0.

    So A = 1, B = 2, C = 3 and D = 6.

    However this is supplementary to the answer as the question only asks for D so answer is D is 6.

    I am gutted that i figured it out in no time at all and was excited to be first to have the answer but alas was stripped of this accolade.

    If A = 1, B = 2, C = 3 and D = 6, that implies 3*1 + 3*2 = 3*6. Since 9 ≠ 18, the only solution which satisfies the equation and the clues is A=2, B=4, C=3, D=6 which was pointed out several times already. You can check it on WolframAlpha.


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  • #42
    DOC09UNAM
    Closed Accounts Posts: 8,305 ✭✭✭DOC09UNAM


    Adyx wrote: »
    If A = 1, B = 2, C = 3 and D = 6, that implies 3*1 + 3*2 = 3*6. Since 9 ≠ 18, the only solution which satisfies the equation and the clues is A=2, B=4, C=3, D=6 which was pointed out several times already. You can check it on WolframAlpha.

    Your not multiplying them at all. You are told the letters stand for digits, not numbers.


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