Results 1 to 10 of 10

Math Help - Solve using proofs/justification

  1. #1
    Junior Member
    Joined
    Dec 2010
    Posts
    61

    Solve using proofs/justification

    Hi, I have a question that needs to be solved using "clear, concise and accurate mathematical justification/proofs." I am having some difficulty on understanding where to start and how to solve the problem. Any help will be appreciated. Below is the question:

    1) You are given four digits (all strictly positive, i.e. greater than 0)
    arranged in a square with a distinct digit from 1 to 9 in each quarter. For example:

    5 6
    3 2

    You make this arrangement into 4 integers by reading horizontally and vertically
    and then adding these numbers up, to get the sum S. So, in the above example,

    S = 56 + 32 + 53 + 62 = 203

    In how many different ways can you fill in such a square so that S = 200?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: Solve using proofs/justification

    Quote Originally Posted by NFS1 View Post
    Hi, I have a question that needs to be solved using "clear, concise and accurate mathematical justification/proofs." I am having some difficulty on understanding where to start and how to solve the problem. Any help will be appreciated. Below is the question:

    1) You are given four digits (all strictly positive, i.e. greater than 0)
    arranged in a square with a distinct digit from 1 to 9 in each quarter. For example:

    5 6
    3 2

    You make this arrangement into 4 integers by reading horizontally and vertically
    and then adding these numbers up, to get the sum S. So, in the above example,

    S = 56 + 32 + 53 + 62 = 203

    In how many different ways can you fill in such a square so that S = 200?
    I won't do the question for you, but here's how you might start. Suppose that the four distinct digits are a,b,c,d, arranged in the form \textstyle{a\atop c}\ {b\atop d}. The four numbers to be added are 10a+b, 10 c+d, 10a+c and 10b+d. Adding them up, we get the equation 20a+11(b+c)+2d=200. For the left side to be an even number, b+c must be even, say b+c=2k. Then the equation becomes 10a+11k+d=100.

    For each possible value of a from 1 to 8 ( a cannot be 9) there is exactly one possible pair of values for k and d. So you can make a table, like this:

    . . . . . . . . . . \begin{array}{c|cccccccc} a&1&2&3&4&5&6&7&8 \\ k&8 \\ d&2 \end{array}

    (I have just filled in the first row and column). For each column in the table, you then have to see how many ways you can choose b and c so that b+c=2k (remembering each time that the four digits a,b,c,d must all be distinct). Having done that, you can just count up the total number of solutions.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539

    Re: Solve using proofs/justification

    Hello, NFS1!

    1) You are given four digits (all strictly positive, i.e. greater than 0)
    arranged in a square with a distinct digit from 1 to 9 in each quarter.

    For example: . \begin{bmatrix}5&6 \\ 3& 2\end{bmatrix}

    You make this arrangement into 4 integers by reading horizontally
    and vertically and then adding these numbers to get the sum S.

    So, in the above example: . S \:\:= 56 + 32 + 53 + 62 \:=\: 203

    In how many different ways can you fill in such a square so that S = 200 ?

    We have the array: . \begin{bmatrix}a&b \\ c&d \end{bmatrix}

    The two-digit numbers are: . 10a+b,\;10c+d,\;10a+c,\;10b+d

    Their sum is 200: . (10a+b) + (10c+d) + (10a+c) + (10b+d) \:=\:200

    . . 20a + 11b + 11c + 2d \:=\:200 \quad\Rightarrow\quad 11b + 11c \:=\:200 - 20a - 2d

    . . 11(b+c) \:=\:2(100 - 10a - d) \quad\Rightarrow\quad b+c \:=\:\frac{2(100-10a-d)}{11}

    We have: . {\color{blue}b + c \:=\:\frac{2\big[100 - (10a + d)\big]}{11}}


    10a+d is a two-digit number
    . . and 100 - (10a+d) must be divisible by 11.

    There are only 8 cases to consider.


    10a+d \,=\,12\!:\;b+c \:=\:\tfrac{2(88)}{11} \:=\:16

    (b,c) \:=\:(7,9),\,\rlap{////}(8,8),\,(9,7) . . . 2 ways


    10a+d \,=\,23\!:\;b+c \:=\:\tfrac{2(77)}{11} \:=\:14

    (b,c) \:=\:(5,9),\,(6,8),\,\rlap{////}(7,7),\,(8,6),(9,5) . . . 4 ways


    10a + d \,=\,34\!:\;b+c \:=\:\tfrac{2(66)}{11} \:=\:12

    (b,c) \:=\:\rlap{////}(3,9),\,\rlap{////}(4,8),\,(5,7),\,\rlap{////}(6,6),\,(7,5),\,\rlap{////}(8,4),\,\rlap{////}(9,3) . . . 2 ways


    10a+d \,=\,45\!:\;b+c \:=\:\frac{2(55)}{11} \:=\:10

    (b,c) \:=\:(1,9),\,(2,8),\,(3,7),\,\rlap{////}(4,6),\,\rlap{////}(5,5),\,\rlap{////}(6,4),\,(7,3),\,(8,2),\,(9,1) . . . 6 ways


    10a+d\,=\,56\!:\;b+c \:=\:\tfrac{2(44)}{11} \:=\:8

    (b,c) \:=\:(1,7),\,\rlap{////}(2,6),\,\rlap{////}(3,5),\,\rlap{////}(4,4),\,\rlap{////}(5,4),\,\rlap{////}(6,2),\,(7,1) . . . 2 ways


    10a+d \,=\,67\!:\;b+c \:=\:\tfrac{2(33)}{11} \:=\:6

    (b,c) \:=\:(1,5),\,(2,4),\,\rlap{////}(3,3),\,(4,2),\,(5,1) . . . 4 ways


    10a + d \,=\,78\!:\;(b+c \:=\:\tfrac{2(22)}{11} \:=\:4

    (b,c) \:=\:(1,3),\,\rlap{////}(2,2),\,(3,1) . . . 2 ways


    10a+d \,=\,89\!:\;b+c \:=\:\tfrac{2(11)}{11} \:=\:2

    (b,c) \:=\:\rlap{////}(1,1) . . . 0 ways


    Hence, there are: . 2 + 4 + 2 + 6 + 2 + 4 + 2 \:=\:22\text{ ways.}


    \text{But in each of these arrays, }a\text{ and }d\text{ can be interchanged.}

    \text{Therefore, there are: }\:2 \times 22 \:=\:44\text{ ways.}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: Solve using proofs/justification

    Quote Originally Posted by Soroban View Post
    \text{But in each of these arrays, }a\text{ and }d\text{ can be interchanged.}
    No they can't.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591

    Re: Solve using proofs/justification

    both Opalg's and Soroban's method both give that a=8 is not possible. although their reasoning in finding b+c differ, both give the same possible pairs (b,c).

    so, given a in {1,2,3,4,5,6,7,8}, if 10a+d+11(b+c)/2 = 100, there can be only one possible choice for (b+c)/2 and d.

    for example, if a = 3, then 70 - d is divisible by 11. since d < 10 (in fact, it's actually < 9), (b+c)/2 has to be 6, so b+c = 12, and from

    70 - d = 66, we get d = 4.

    if we choose a = 4, we do NOT get d = 3. if a = 4, then 60 - d is divisible by 11, so (b+c)/2 = 5, hence b+c = 10. from 60 - d = 55, we get d = 5.

    so each pair (a,d) is uniquely defined, we need only count them once.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2010
    Posts
    61

    Re: Solve using proofs/justification

    Quote Originally Posted by Deveno View Post
    both Opalg's and Soroban's method both give that a=8 is not possible. although their reasoning in finding b+c differ, both give the same possible pairs (b,c).

    so, given a in {1,2,3,4,5,6,7,8}, if 10a+d+11(b+c)/2 = 100, there can be only one possible choice for (b+c)/2 and d.

    for example, if a = 3, then 70 - d is divisible by 11. since d < 10 (in fact, it's actually < 9), (b+c)/2 has to be 6, so b+c = 12, and from

    70 - d = 66, we get d = 4.

    if we choose a = 4, we do NOT get d = 3. if a = 4, then 60 - d is divisible by 11, so (b+c)/2 = 5, hence b+c = 10. from 60 - d = 55, we get d = 5.

    so each pair (a,d) is uniquely defined, we need only count them once.
    so they are only 22 pairs??
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Dec 2010
    Posts
    61

    Re: Solve using proofs/justification

    I also have a second part to the problem that i could also do with some help with:

    B) In the above example S = 203. Clearly
    83 <= S <= 357: (1) (less than or equal to)
    (if you allow repetitions then 44 <= S <= 396) For how many values of S satisfying
    the inequality (1) can a square not be constructed. For example, it is not possible
    to construct a square with S = 4
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: Solve using proofs/justification

    Quote Originally Posted by NFS1 View Post
    I also have a second part to the problem that i could also do with some help with:

    B) In the above example S = 203. Clearly
    83 \leqslant S \leqslant 357\qquad (1)
    (if you allow repetitions then 44 <= S <= 396) For how many values of S satisfying
    the inequality (1) can a square not be constructed. For example, it is not possible
    to construct a square with S = 4
    I don't see any easy way to do this. You have to go back to the equation 20a+11(b+c)+2d = S and check (for each value of S) whether there is a solution in which a, b, c and d are distinct digits. The smallest possible value of S is 83 (obtained when a=1, b=2, c=3, d=4) and the largest possible value is 357 (a=9, b=8, c=7, d=6). The only values of S for which no solution is possible are going to occur when S is fairly close to those extreme values. When S is well inside the interval [83,357] (for example, when S=200, as in the original problem) there is enough flexibility in the choice of a,b,c,d to ensure that there will always be at least one solution.

    Notice also that if (a,b,c,d) is a solution for S, then (10a,10b,10c,10d) will be a solution for 440S. So the values of S for which there is no solution will come in pairs adding up to 440. I think that there are 22 such values, namely

    . . . . . . . . . . \begin{array}{cccc}84&86&88&90 \\ 95&97&99 \\ 106&108&110 \\ 117&&&323 \\ &330&332&334 \\ &341&343&345 \\ 350&352&354&356\rlap{.} \end{array}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Dec 2010
    Posts
    61

    Re: Solve using proofs/justification

    Quote Originally Posted by Opalg View Post
    I don't see any easy way to do this. You have to go back to the equation 20a+11(b+c)+2d = S and check (for each value of S) whether there is a solution in which a, b, c and d are distinct digits. The smallest possible value of S is 83 (obtained when a=1, b=2, c=3, d=4) and the largest possible value is 357 (a=9, b=8, c=7, d=6). The only values of S for which no solution is possible are going to occur when S is fairly close to those extreme values. When S is well inside the interval [83,357] (for example, when S=200, as in the original problem) there is enough flexibility in the choice of a,b,c,d to ensure that there will always be at least one solution.

    Notice also that if (a,b,c,d) is a solution for S, then (10a,10b,10c,10d) will be a solution for 440S. So the values of S for which there is no solution will come in pairs adding up to 440. I think that there are 22 such values, namely

    . . . . . . . . . . \begin{array}{cccc}84&86&88&90 \\ 95&97&99 \\ 106&108&110 \\ 117&&&323 \\ &330&332&334 \\ &341&343&345 \\ 350&352&354&356\rlap{.} \end{array}
    Thanks for the reply, but how did you reach the 22 values using "clear, concise and accurate mathematical justification/proofs." Sorry for being a pain
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: Solve using proofs/justification

    Quote Originally Posted by NFS1 View Post
    how did you reach the 22 values using "clear, concise and accurate mathematical justification/proofs."
    I didn't. That suggested solution is just a mixture of trial and error, and guesswork. I was hoping that someone else would confirm or correct it, and hopefully indicate a more convincing method.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. power series justification
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 11th 2011, 10:52 AM
  2. Simple linear algebra justification - clarification
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 6th 2010, 08:00 PM
  3. Replies: 0
    Last Post: April 28th 2009, 06:09 AM
  4. Determining the confidence interval and claim justification
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: March 17th 2009, 05:59 AM
  5. [SOLVED] Question about justification (pdf)
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: February 7th 2009, 07:01 AM

Search Tags


/mathhelpforum @mathhelpforum