Problem of the Month: Courtney's Collection

[Jpknox]

My gifted students work on Problems of the Month from our district, but they come with no solutions... help!

Level C
Courtney visited her grandmother. Her grandmother used to collect stamps. She had a shoebox full of 5 cent, 6 cent, and 7 cent stamps. Courtney thought, "I could make a lot of different-sized letters and postcards with these stamps." She tried to figure out the different amount of postage she could make with a combination of those stamps. What totals could she make what totals are impossible? Explain how you found your solution. How do you know your solution is correct?

Level D
Courtney's grandmother said, "Your grandpa has different shoeboxes with other stamps. His shoebox has just 4 cent, 6 cent, and 8 cent stamps. Which totals can you make with these stamps?"

Courtney said, "I wonder why there is a different between Grandma's and Grandpa's shoeboxes. I like finding the largest total I can't make, but Grandpa's box has an unlimited amount of totals I can't make. I wonder why it works for some and not for others." She continues, "I am going to investigate which three stamp amounts are like Grandma's shoebox and which three stamp amounts are like Grandpa's shoebox. I am going to compare three different sets. I will try 6 cents, 7 cents, and 8 cents. Then I will try 6 cents, 9 cents, and 12 cents. Finally I will try 6 cents, 8 cents, and 9 cents." Explain how the three different sets are like either Grandma's shoebox or Grandpa's shoebox.

Level E
Determine a method for predicting whether a given set of any three stamps would have a largest impossible total, or infinite impossible totals. Justify your method using mathematical reasoning.

For those sets of three stamps that have a largest impossible total, make a conjecture about how to find that total.

Help.

 

[stapel]

My gifted students work on Problems of the Month from our district, but they come with no solutions... help!

You're the teacher. With what are you asking for help? :shock:

 

[Ishuda]

Just some thoughts on the problem:

Let S_n be the solution set of possible totals for n number of denominations. Then the set of impossible solutions is I - S_n.

Let X={x_j} be the set of the number of stamps of denomination y_j where y_j belongs to Y which is the ordered set {y_j such that y_j < y_k if j<k}. For example, in Level C, n = 3, y₁ = 5, y₂ = 6, y₃ = 7, and we are not given x₁, x₂, or x₃. The largest total one could make would
T = x₁* y₁ + x₂ * y₂ + x₃ * y₃ + ... + x_n* y_n
Therefore all totals larger than T can not made from a combination of the given stamps. So, unless you allow numbers larger than T, there are always a finite number of totals which can't be made. The converse of that is to let some x be infinite (should really be all x's for the problems to make sense). So all x's infinite is what is assumed.

We can assume that for j < k, j = 1, 2, 3, ..., n there is no y_k = F y_j since, if not, the solution set S_n can be reduced set S_n-1. That is, assume there are two y's, y_i and y_k with j < k and F>1 such that y_k = F y_k. Then
t = a₁* y₁ + ... + a_j * y_j + ... + a_k * y_k + ... + a_n* y_n
= a₁* y₁ + ... + a_j * y_j + ... + a_k * F * y_j + ... + a_n* y_n
= a₁* y₁ + ... (a_j + a_k * F) * y_j + ... + a_k-1 y_k-1 + a_k+1 y_k+1 +... + a_n* y_n
where (a_j + a_k * F) can take on any non-negative value. Thus S_n= S_n-1= S_n-2 = ... = S_m where S_m is the first set in the list which satisfies our condition of no y is a (greater than 1) multiple of any other y. This also means the grandfathers set of stamps is equivalent to a set containing only 4 cent and 6 cent stamps.

Obviously any set of stamps which contains the 1 cent stamp can make every total, so we can assume y>1. Also every set S_n has at least the one member 0, i.e. we can choose to use 0 stamps and set T_n is not empty since it does not contain one [remember we are restricting our self to set with y>1]. We should also note, to be general, that it is not the set S_n since there may be many such sets with n elements, _mC_n to be exact where m is the total number of denominations allowed [however all of these may not satisfy our no y is a multiple of different y criterion].

Sorry, ran out of time. Be interested to see other posts.

 

[Jpknox]

You're the teacher. With what are you asking for help? :shock:

How incredibly helpful! Thank goodness I posted this online!

 

[Jpknox]

Just some thoughts on the problem:

Let S_n be the solution set of possible totals for n number of denominations. Then the set of impossible solutions is I - S_n.

Let X={x_j} be the set of the number of stamps of denomination y_j where y_j belongs to Y which is the ordered set {y_j such that y_j < y_k if j<k}. For example, in Level C, n = 3, y₁ = 5, y₂ = 6, y₃ = 7, and we are not given x₁, x₂, or x₃. The largest total one could make would
T = x₁* y₁ + x₂ * y₂ + x₃ * y₃ + ... + x_n* y_n
Therefore all totals larger than T can not made from a combination of the given stamps. So, unless you allow numbers larger than T, there are always a finite number of totals which can't be made. The converse of that is to let some x be infinite (should really be all x's for the problems to make sense). So all x's infinite is what is assumed.

We can assume that for j < k, j = 1, 2, 3, ..., n there is no y_k = F y_j since, if not, the solution set S_n can be reduced set S_n-1. That is, assume there are two y's, y_i and y_k with j < k and F>1 such that y_k = F y_k. Then
t = a₁* y₁ + ... + a_j * y_j + ... + a_k * y_k + ... + a_n* y_n
= a₁* y₁ + ... + a_j * y_j + ... + a_k * F * y_j + ... + a_n* y_n
= a₁* y₁ + ... (a_j + a_k * F) * y_j + ... + a_k-1 y_k-1 + a_k+1 y_k+1 +... + a_n* y_n
where (a_j + a_k * F) can take on any non-negative value. Thus S_n= S_n-1= S_n-2 = ... = S_m where S_m is the first set in the list which satisfies our condition of no y is a (greater than 1) multiple of any other y. This also means the grandfathers set of stamps is equivalent to a set containing only 4 cent and 6 cent stamps.

Obviously any set of stamps which contains the 1 cent stamp can make every total, so we can assume y>1. Also every set S_n has at least the one member 0, i.e. we can choose to use 0 stamps and set T_n is not empty since it does not contain one [remember we are restricting our self to set with y>1]. We should also note, to be general, that it is not the set S_n since there may be many such sets with n elements, _mC_n to be exact where m is the total number of denominations allowed [however all of these may not satisfy our no y is a multiple of different y criterion].

Sorry, ran out of time. Be interested to see other posts.

Thank you so much for your help! I probably should have added that my students are in 3rd grade. Our district wants us to be using Problems of the Month where we focus on the path and process to the solution rather than just the solution so they purposely do not give us solutions. However, gifted students want things explained to them thoroughly and explicitly and I just didn't know how to do that with these problems! Level A and B were fine - the difficulty of the problems range from K-12 where Level A is K-2, Level B is 3-5, Level C is 6-8, Level D is 9-10, and Level E is 11-12+. I am suppose to explain this to them since the month is finished, but was trying to find a way to do it in easier terms since they are unfamiliar with higher level math at this age.

However, thank you so much for clarifying for me! I really appreciate it!

 

[Ishuda]

Thank you so much for your help! ...

You are certainly welcome. One of my failings is that, in a lot of situations, I was never much good explaining things 'in English', math had to get involved some way so I'm impressed by people who can do that, especially to those who have little or no math. So, good luck.

BTW: another thought about the problem; if there is a common multiple M>1 in the set of stamps, such as in grandfathers collection of 4 and 6 cent stamps with M = 2, then there is always an infinite number of impossibles, i.e. if there is such an M, then yj = M uj and
t = a₁* y₁ + ... + a_j * y_j + ... + a_k * y_k + ... + a_n* y_n
= M (a₁* u₁ + ... + a_j * u_j + ... + a_k * u_k + ... + a_n* u_n)
= M L
and all totals in S_n are multiples of M. Consider the set of totals with a₁ = j and the remainder of a's equal to zero. The totals set for this set of a's is {j M u₁, j = 1, 2, 3, ...} which is a subset of S_n and the set {j M u₁ - 1, j = 1, 2, 3, ...} does not belongs to S since they are not multiples of M. The elements are also all positive and there are an infinite number of them. Therefore the set T_n contains an infinite number of elements.





There is always an infinite number of possibles, i.e. j y₁ belongs to S_n for all j.

 

[stapel]

You're the teacher. With what are you asking for help?

...my students are in 3rd grade.... I am suppose[d] to explain th[ese] to them since the month is finished, but was trying to find a way to do it in easier terms....

So you do have a solution, but are wanting help in re-phrasing...? If so, kindly please post your complete solution, specifying the particular portions which you feel could benefit from simplified language. Thank you.