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I'm about to really upset here.. but we have to find where the shrub can go per n, not just whether there's one solution.. so far I've come up with these..

http://www.filelodge.com/files/room21/537756/Positions.xls

I did accept n=2 as a solution, (the first of a few noticed pros of the l-shaped ones over the others) but I can't say I'd really noticed the ability to make the larger Ls from it..

However even when I copy the image to HD and enlarge it i can't really see anything.. (e.g. not a lack of understanding.. problem with the image!)

**Ansette**- Replies: 20

Ok this may be hard to explain..

Imagine a square made up of 1x1 squares, of which for a square of length n, there are n² of these squares..

This is our "garden".

Now.. one of these squares has to be for a shrub..

We're given 3 different types of paving slabs.. first of an L shaped one (e.g. 3 squares, cant really explain other than that), then a 1x3 and finally a 1x4.. we have to work out for what n these can work (having started off with the L's.. the "company" then suggests the next options)

The shrub I must add has to be able to be present in any of the 1x1 squares..

Now, we've proved what n will definitely not work using the following ideas:

The square of side n must fit the pattern (n²-1)/3 ∈ Z+

Now, n can be written as or n=3m+p, where p=0,1 or 2 and m=0,1,2

Thus, n² = 9m² +6mp + p²

And, (n²-1)/3 = (9m² +6mp + p² -1)/3

= 3m² +2mp + (p² -1)/3

The co-efficients for both m² and mp are divisible by 3 so thus whether the whole quadratic is divisible by 3 is dependant on this p² -1 term.

Given the values of p earlier for any n, p²=0,1 or 4. And obviously (p² -1)/3 therefore equals either -⅓, 0 or 1. Thus the pattern will only work for squares that are not divisible by 3.

This would work for both L and 1x3 logically as they are both composed of 3 squares.. and a similiar idea would be used for 1x4..

The thing we want to know is that how can we prove that for any n (other than those discounted above) they will work for the L-shaped blocks (which seemingly we've found simply by trial)..

Also the 1x3 and 1x4 are both worse of in terms of their use.. but how can we show this mathematically.. i presume between the Ls and 1x3s that it's due to the fact that the L has a rotational symmetry order of 1 as opposed to the 1x3s order of 2.. thus meaning it can create more shapes when put with others..but again this is all worded and not particuarly mathematical...

Help me!!

if this is alevel im presuming it's edexcel.. there are a lot of mistakes in the answers bit lol

must be a mistake in the book...

the equation u have is indeed correct..

im curious as to why ur looking at (y-2)²...?

the x²-2x can be simplified to x(x-2)

so:

x(x-2) +y²-4y=4

what it seems you're doing is saying (y-2)² = y²-4y+4.. which is true..

however what you have in this equation is:

x(x-2) = -y²+4y+4

which using the quadratic formula forms:

y=1±√2

so you have x(x-2) = (y-[1+√2])(y-[1-√2])

or aternatively:

x(x-2) = (y-[1+√2])(y+[√2-1])

well so far we have (from n=1)

1,2,1,4,5..

but then we're not even certain these are right!

group of students for this work..

ok im very impressed lol.. i had some trouble making the formula work at first.. misread the last j as a 3 so was writing iC3 each time.. really math it up lol..

there is a third question really..

we have to try and place these individual boxes into a large parcel that must be rectangular with no free space, though the dimensions are irrelevant.. kind of like a three-dimensional tetris..

any thoughts?

(alternatively if one parcel is not possible, as few as possible..)

obviously for n=1, 1 package

for n=2 it requires 2, but then that's understandable with a volume of 15.

for n=3 i believe we've managed 1, but after that we're unsure

ah of course.. working like the binomial theorem..!!

i have to admit i'm slightly confused..

**Ansette**- Replies: 33

We've been givin a problem involving boxes (parcels basically)..

The dimensions of each parcel are integers abc with 1≤a≤b≤c≤n

Basically such that for each n, there are many combinations as you can see, for instance with n=2, the options are:

1,1,1

1,1,2

1,2,2

2,2,2

to fit within those bounds.

The first thing we've noticed is that the number of parcels per each n is a tetrahedral number.. e.g. general formula with our n = n(n+1)(n+2)/6

e.g. 1,4,10,20,35.. etc

The next thing of note is that the total volume for each set of parcels fits the pattern of is 1,15,90,350,1050.. Stirling numbers of second order so i'm informed by the internet..

What I'm trying to ask is for an explanation as to why this is the case?

(And also a general formula for stirling numbers of second order if possible!)

well.. a quadratic function is generally given in the form:

ax²+bx+c=0

where a,b,c are real constants.

The value of x is the point on the x-axis that the quadratic function touches it, if it touches once then there's one solution however it can cross it then come back up as it's a curve and thus there are 2 solutions.

Now, you know there's only one solution at (3,0).. e.g. x=3.

therefore the family logically is:

a(3²)+b(3)+c=0

therefore:

9a+3b+c=0

(or alternatively as c is whats called a dummy variable, you could simply write it using a new value and simplify it)

3a+b=d

hope this makes sense, i am very tired lol :s

no, see the dimensions must be 1,1,1 or 1,1,2 or 1,2,2 or 2,2,2

because 1≤a≤b≤c≤n

that's 4 different combinations. however what im now querying of the lecturer is that for each n it's not one of the combinations, there are indeed all of those packages (so for n=2, 4 packages.)

any idea on why the combinations take those patterns? i mean i can see it but unless i can prove it.. it's no use to us :s

so for combinations, forn any n.

Number of packages = (n)(n+1)(n+2)/6 or alternative (n+2)C3 [(n+2)Choose 3]

In the case that there are x many packets with the dimensions bounded as suggested above...

I've been looking at the number of packets there are.. e.g. each x for n.

n=2, x=4

n=3, x=10

n=4, x=20

n=5, x=35

This sequence by the pyramidal numbers...

Another thing I noticed, the number of occurences of each size per set.

In other words, for instance with n=2,

out of all the combinations, the number 1 appears 6 times, the number 2 also 6 times.

for each n, the number of occurences is the same (for that n).

the number of occurences (y) is as follows:

n=2, y=6

n=3, y=10

n=4, y=15

n=5, y=21

This sequence being the triangular numbers..

Can anyone give me a reason as to why this is the case?

Thats the thing.. we really do not understand! That is literally what we've been given..

**Ansette**- Replies: 7

Hey,we've been set an assignment in pure maths and have no clue how to go about it..

"A student on a year abroad buys a number of presents and wants to send them home. The goods are already wrapped up as individual packets: by a curious chance there is an integer n such that there is precisely one packet of depth a, width b and length c, for each set of integers (a,b,c) with 1<=a<=b<=c<=n

The student could send each packet home separately, but it is cheaper (and mathematically more interesting) to make the packets up into parcels. To avoid breakages, the parcels must have no empty spaces (the country is experiencing a shortage of bubble wrap). In other words, a parcel of size pxqxr will have volume pqr that is equal to the sum of the packets it contains. Assuming that it does not matter how large the parcels are, how many parcels does the student need?"

Seriously we're so unsure what to do :s

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**Ansette**- Replies: 13

We have been given the problem of trying to find the minimum surface area given two variables in such a shape as the one in the image uploaded. any ideas?

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