Results 1 to 10 of 10

Thread: 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
    10

    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 $\displaystyle a,b,c,d$, arranged in the form $\displaystyle \textstyle{a\atop c}\ {b\atop d}$. The four numbers to be added are $\displaystyle 10a+b$, $\displaystyle 10 c+d$, $\displaystyle 10a+c$ and $\displaystyle 10b+d.$ Adding them up, we get the equation $\displaystyle 20a+11(b+c)+2d=200.$ For the left side to be an even number, $\displaystyle b+c$ must be even, say $\displaystyle b+c=2k$. Then the equation becomes $\displaystyle 10a+11k+d=100.$

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

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

    (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 $\displaystyle b$ and $\displaystyle c$ so that $\displaystyle b+c=2k$ (remembering each time that the four digits $\displaystyle a,b,c,d$ 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
    12,028
    Thanks
    848

    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: .$\displaystyle \begin{bmatrix}5&6 \\ 3& 2\end{bmatrix}$

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

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

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

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

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

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

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

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

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


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

    There are only 8 cases to consider.


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

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


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

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


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

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


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

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


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

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


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

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


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

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


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

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


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


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

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

    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
    10

    Re: Solve using proofs/justification

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

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    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
    10

    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
    $\displaystyle 83 \leqslant S \leqslant 357\qquad (1)$
    (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 $\displaystyle 20a+11(b+c)+2d = S$ 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

    . . . . . . . . . .$\displaystyle \begin{array}{cccc}84&86&88&90 \\ 95&97&99 \\ 106&108&110 \\ 117&&&323 \\ &330&332&334 \\ &341&343&345 \\ 350&352&354&356\rlap{.} \end{array}$
    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 $\displaystyle 20a+11(b+c)+2d = S$ 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

    . . . . . . . . . .$\displaystyle \begin{array}{cccc}84&86&88&90 \\ 95&97&99 \\ 106&108&110 \\ 117&&&323 \\ &330&332&334 \\ &341&343&345 \\ 350&352&354&356\rlap{.} \end{array}$
    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
    10

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

Search Tags


/mathhelpforum @mathhelpforum