Math Is Fun Forum

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

I don't understand the question!

Here is the question. I don't get part(B) at all. Doesn't make any sense to me. Can someone explains it a little bit please. Where did 83 and 357 come from (or 44 and 396)? I don't understand what I'm asked to do. Please help!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: I don't understand the question!

hi bubokribuck

It says distinct digits which I think means there are no repeats.  So the smallest total would be to use 1, 2, 3, and 4 giving a total of

and the largest would use 9, 8, 7 and 6
giving a total of

If repetitions are allowed then use 1, 1, 1, and 1 for the smallest and 9, 9, 9, 9 for the largest.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Thank you Bob! Now it makes more sense. However, I'm still not sure what the question is asking though. Is it asking to find values of S where it's impossible like S=47? If so, how am I supposed to do this? (I can do part A because the value of S is given, but part(B) seems different and I'm not confident with it).

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

hi bubokribuck;

Here is the question. I don't get part(B) at all

Did you get the answers?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Did you get the answers?

Hi, I've worked out the answer for part(A), but doesn't have a clue about part(B), still trying to figure out how to make a start
(I haven't been given the answers by the way)

Last edited by bubokribuck (2011-12-17 04:32:53)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Hi;

Did you get all the solutions?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Hi;

Did you get all the solutions?

Do you mean if I got the values for S where a square cannot be constructed? If so, then no I didn't get the solutions.
I was just given the question as it is, and was asked to solve it on my own, but I'm a bit confused with the question itself, therefore I posted here for help

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Hi;

Okay, but there are 22 solutions for S = 200 so be careful. They did say they want all the arrangements.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Hi;

Okay, but there are 22 solutions for S = 200 so be careful. They did say they want all the arrangements.

Yes I got 22 for part(A), thanks.

But I'm stuck at (B) at the moment , any help please?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

They want you with digits repeated or not to find all the squares that can not be constructed. If your method was good for the other S = 200 then it should apply here. Or I can run them off.

For instance even using repetitive digits 47 can not be made with that square.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

They want you with digits repeated or not to find all the squares that can not be constructed. If your method was good for the other S = 200 then it should apply here. Or I can run them off.

For instance even using repetitive digits 47 can not be made with that square.

So I need to set S to all the numbers range from 83 all the way to 357 and solve respectively!?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Hi;

From 44 to 396.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Hi;

From 44 to 396.

But the questions says "distinct digits" though.
But either case, does that mean I need to apply the method I used in (A) but this time for 83≤S≤357 (or 44≤S≤396)? It looks horribly time consuming, and our lecturer said that "Don’t attempt to answer these questions by trial and error it will take you a very long time!".....

Would you give me a hint about how to make a start please, as I'm really confusing at the moment.
Thanks!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Yes, it does so you will have to go from 83 to 357.

What kind of class is this?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Yes, it does so you will have to go from 83 to 357.

What kind of class is this?

I'm doing a maths degree in uni
Thanks for the help!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Hi;

What was the method that you have been using for problems of this type?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Hi;

What was the method that you have been using for problems of this type?

I didn't learn any method for solving such problem. Our lecturer set it as our coursework and asked us to do our own research.

Basically I set the square as
a b
c d

So if S=200, 200-20a-2d=11(b+c) where 200-20a-2d is divisible by 11. I listed all the possible values for a and d, once all listed, I moved on to find the values for b and c.

This is the solution I've come up with for part (A) of the question, which is why I think doesn't apply to (B) as then I need to do the method for 274 times which sounds really scary

Last edited by bubokribuck (2011-12-17 11:37:09)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Hi;

Yes, the work will be quite difficult. This is an assignment more for a computer class.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: I don't understand the question!

hi bubokribuck

I do not know if the following will be helpful but here goes:

You have obviously got to

Now write this as

Now treat b + c as a single number.

It can only have values from 3 (= 1 + 2) up to 17 (= 8 + 9) so will contribute part totals of 33, 44, 55 ... 187.

b and c cannot take the same values during this analysis which makes life easier.

Similarly, 10a + d can only have values from 12 to 98, giving part totals of 24, ... 196.

You will have to remove the cases where a = d from this.

And then, the hard bit I think.  Remove the cases where a or d has a value already used by b or c.

It looks like it will be easier to consider the 'repeats allowed' problem first.

Hope that helps.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

hi bubokribuck

I do not know if the following will be helpful but here goes:

You have obviously got to

Now write this as

Now treat b + c as a single number.

It can only have values from 3 (= 1 + 2) up to 17 (= 8 + 9) so will contribute part totals of 33, 44, 55 ... 187.

b and c cannot take the same values during this analysis which makes life easier.

Similarly, 10a + d can only have values from 12 to 98, giving part totals of 24, ... 196.

You will have to remove the cases where a = d from this.

And then, the hard bit I think.  Remove the cases where a or d has a value already used by b or c.

It looks like it will be easier to consider the 'repeats allowed' problem first.

Hope that helps.

Bob

Hi Bob, since your suggestion, I've come up with quite a lot attempts but every time failed to reach a correct conclusion. The following is what I think I've done right so far.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Hi bubokribuck;

Are you making any progress yet? Here is a little hint that will allow you to handle it in a totally mechanical way.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Hi bubokribuck;

Are you making any progress yet? Here is a little hint that will allow you to handle it in a totally mechanical way.

Hi Bob, thanks for your reply. I'm still working on it, but looks like I've done it wrong again. I've asked my tutor and he said that we don't need to take the repetition into account. So does that affect the answer?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: I don't understand the question!

Hi bubokribuck;

It sure does and it makes the problem harder to do by the generating function approach.
I am still working on it.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

bubokribuck

Member

Registered: 2011-10-28

Posts: 42

Re: I don't understand the question!

Hi bubokribuck;

It sure does and it makes the problem harder to do by the generating function approach.
I am still working on it.

Thanks, I'm working on it too. It looks like an endless nightmare