Problems and Solutions

mathsyperson · 2005-08-02 20:31:49

Jai Ganesh · 2005-08-02 21:00:34

Mathsy is right This is the complete solution.
Assuming my normal speed is 's' and the normal time taken is 't',
Distance/Time = Speed (or) Distance/Speed = Time
We get 840/s = t ---------------------(1)
The second equation would be 840/(s+5) = t-3
840 = (t-3)(s+5)
840 = st+5t-3s-15 -------------(2)
Subtracting (1) from (2) 840 = st we get
0 = 5t-3s-15
Putting t = 840/s (from (1), 0 = 5(840/s) -3s - 15
0 = 4200/s - 3s - 15
0 = 4200 - 3s² -15s
3s² +15s - 4200 = 0
Dividing by 3 s² + 5s - 1400 = 0
s = [-5 ± √(25+5600)]/2
s = [-5 ± √(5625)]/2
s = (-5 ± 75 )/2
Taking the positive value, s = 35
Therefore, my normal speed is 35 kilometres/hour.

Jai Ganesh · 2005-08-03 16:05:30

Problem # n+1
We know water absorbs light by the fact that it is dark in the ocean depths. If 10 metres of deep water absorbs 40% of the light which strikes the surface, at what depth would light striking the surface be reduced to 1/300,000 th? Assume that the rate of absorption is proportional to the light.

mathsyperson · 2005-08-03 21:04:58

I'm not entirely sure this is right, because I'm not completely comfortable with logs, but I think that it's

Jai Ganesh · 2005-08-04 16:29:22

Great work, Mathsy!
I don't know how you solved the problem, but this is how I would do it.

Let 'l' denote the light that strikes the surface of water at depth 'h' then
dl/dh = -kl
dl/l = -kdh
Integrating,
log l = -kh + c
Let log l = lo when h=0, so that
log (l/lo) = -kh
l/lo = (100-40)/100 = 3/5 when h=10
therefore, log 3/5 = -10k --------------------------(I)
We have to find h when
l/lo = 1/300,000
log (1/300,000) = -hk -----------------------------(II)
Dividing II by I,
-log 300,000/ log 0.6 = h/10,
h=247 metres approximately.

Jai Ganesh · 2005-08-04 19:04:54

Problem #n+2

Divide $1,000 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $1,000 and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?

MathsIsFun · 2005-08-04 19:35:05

You will need a bag with $1 in it.

Possibly a bag with $2 in it.

And that will mean that if you want $3 I can just give you the $1 bag and the $2 bag.

Possibly the next choice would be a $4 bag ... or maybe another $2 bag ...

MathsIsFun · 2005-08-04 19:44:35

ganesh wrote:

... If 10 metres of deep water absorbs 40% of the light which strikes the surface...

So, that means every 10 metres the light is reduced to 60%.

After 20 metres it is further reduced by 60%, leaving 0.6×0.6 = 0.36 (36%)
After 30 metres it is further reduced by 60%, leaving 0.6×0.6×0.6 = 0.6^3 = 0.216 (26%)
...
After 250 metres, there would be 0.6^25 = 0.000002843 (0.00028%)

And 0.000002843 = 1/350,000 th of the original light

Wow, pretty dark, and only 250 metres down.

mathsyperson · 2005-08-05 00:35:51

I can't think of a better system than the one MathsIsFun started. That is, $1, $2, $4, $8 and so on with powers of 2. You can go up to $256 in this way and have 8 bags, but the next one you need is $512 and you don't have enough money left. I'm tempted to say that you can fill the last bag with the remaining amount, $489, but I'm not entirely sure that that would give all the amounts. If this does work, then the answer would be nine bags.

Of course, in reality, you'd only need one bag, because no one would ever ask for anything except $1000.

Jai Ganesh · 2005-08-07 16:44:16

You are right, Mathsy, but the number of bags required is 10!
They would contain $1, $2, $4, $8, $16, $32, $64, $128, $256 and $489. For any amount less than or equal to $511, the first 9 bags can be used. For amount greater than $511, one bag of $489 and remaining from one or more of the first nine bags can be used.

Jai Ganesh · 2005-08-07 17:09:29

Problem #n+3
A rectangular metal sheet has dimensions 30m, 14m.
Four equal squares are cut off from the four corners of the rectangle.
Then, the sheet is folded to form a cuboidal box with top open.
What is the maximum volume of the cuboid you can get?

mathsyperson · 2005-08-07 19:57:19

Oh, buckets. I was thinking 256=2^8, so there must be 8 bags, forgetting that 2^0 was included. Well, I got the gist.

Back to this puzzle, you're cutting 2 squares from each side, so whatever the length of that square is, the 30 and 14 are reduced by 2 times that length. The height of the cuboid is that length, so there will be an optimum value.
The length would be 30-2x, the width would be 14-2x and the height would be x.
So, y=x(30-2x)(14-x) and we want the maximum value of y.
Multiply out: y=4x³-88x²+420x.
Hmm, maximum point, eh? Completing the square would be quite complicated, so this looks like a job for *drumroll please* CALCULUS!
Differentiating: dy/dx=12x²-176x+420.
The maximum point is so called because it is not getting any higher, but the points before it are not higher than it either. This means that, for that point on the graph, a tangent to it would be horizontal. This means that the gradient would be 0. dy/dx is the gradient, so we need to solve 12x²-176x+420=0. MathsIsFun's solvy tool says that x=3 or 10 2/3, but if it was the latter then the width would be negative, so it is disregarded. So, x=3 and so the length is 24, width is 8 and height is 3, meaning the volume is 576.
So, what happens if we're allowed to cut rectangles?

Jai Ganesh · 2005-08-08 16:33:22

Absolutely right, Mathsy!
What happens if we're allowed to cut rectangles?
We get two variables, x and y.
Volume = (30-2x)(14-2y)(2x)
How do we proceed? Any clues?

Jai Ganesh · 2005-08-08 16:35:07

Problem # n+4
(1) How many diagonals would a polygon of 'n' sides contain?
(2) What would be the sum of the internal angles of a polygon of 'n' sides?

mathsyperson · 2005-08-08 19:10:05

The second one is 180(n-2), unless you have a different definition of internal angle, but I don't know about the first one.
I think there probably is a formula for it, so I'll try to figure them out.
Triangle: 3 sides, 0 diagonals.
Quadrilateral: 4 sides, 2 diagonals.
Pentagon: 5 sides, 5 diagonals.
Hexagon: 6 sides, 9 diagonals.
It seems to be an arithmetic sequence, so a heptagon should have 9+5=14 diagonals.
Heptagon: 7 sides, 14 diagonals, and very starry they are too.
As the difference is 1, it means that you're just adding the triangular numbers to a constant, in this case -1.
The sequence also starts at 3, so that means that d=(n-2)(n-1)/2-1. Both of these formulas (it's a word) apply only when n>2, ∈Z.

About the rectangles in your previous problem, think about it a bit more. I didn't realise the answer straight away either, but trust me, you're barking up the wrong tree at the moment.

Last edited by mathsyperson (2005-08-08 19:10:17)

Jai Ganesh · 2005-08-08 20:24:49

There are n vertices (or points) in a polygon of n sides.
The n points can be connected to any other point in nC2 ways.
That is we get nC2 straight lines. But, of these n straight lines form the sides of the polygon.
Therefore, the number of diagonals is equal to
nC2 - n
= n!/(n-2)!(2!) - n
= n(n-1)/2 - n
= [n(n-1)-2n]/2
= [n² -n-2n]/2 = (n²-3n)/2 = n(n-3)/2

The sum of intenral angles of a polygon of n sides is 180(n-2) degrees as rightly replied by Mathsy!

Jai Ganesh · 2005-08-08 20:31:41

Problme #n+5
How many zeros would 1,000! end in?

Jai Ganesh · 2005-08-08 21:03:36

mathsyperson wrote:

About the rectangles in your previous problem, think about it a bit more.

Got it, Mathsy! By cutting 4 rectangles from the four corners, a cuboidal box cannot be made as the heights of the two adjacent sides would differ!

mathsyperson · 2005-08-08 21:12:28

That's it. I posted that as a genuine question and then realised the flaw a while later.

Your 1000! problem is very interesting. I think it would have

zeroes, but I might have made a mistake somewhere.

Jai Ganesh · 2005-08-08 22:58:38

I get a different answer.
There are 9 numbers ending with zero up to 99. 10,20,30,40,50,60,70,80,90.
100 has two zeros. Then, in each set of ten consecutive numbers (1-10, 11-20, 21-30 etc.),
one ends in 2 and one ends in 5, which gives us another 0. There are ten tens in a hundred.
Therefore, I thought 100! would end with 21 consecuitive zeros.
But when I calculated 100! using the full precision calculator on this website, I got
93,326,215,443,944,152,681,699,238,856,
266,700,490,715,968,264,381,621,468,592,
963,895,217,599,993,229,915,608,941,463,
976,156,518,286,253,697,920,827,223,758,
251,185,210,916,864,000,000,000,000,000,
000,000,000
That is 24 zeros! I wonder where those three extra zeros came from
Okay, one additional zero 40, 50 since 40x50 = 2000, the other by using 4 (instead of 2)for 25 and 8 for 75?

Did I miss something, Mathsy?

Last edited by Jai Ganesh (2005-08-08 23:56:38)

mathsyperson · 2005-08-09 00:09:45

For every 0 on the end of the number, you need a prime factor of 2 and 5. I used Excel's mod function to see how many prime factors of 2 there were in each number from 1 to 1000. eg. 1 has 0, 2 has 1, 3 has 0, 4 has 2 etc.
I did this with =if(mod(a1,2)=0,1,0) and making additional columns, replacing the 2 with 4, 8, 16, 32, 64, 128, 256 and 512. By adding all the values together, I found that 1000!=2^9??*n. I can't remember the exact amount because I didn't save the spreadsheet and it wasn't important because there is a lower power of 5. I did the same thing with 5's except that I used 5, 25, 125 and 625 as the mod values. Adding these gives 249, and so that would be the answer.

There are plenty of extra 2's to choose from in your hundred problem, there are 3 in 16 just to start with, but the 3 extra 5's that you need are in 25, 50 and 75. You took one from each of them, but they each have 2!

Jai Ganesh · 2005-08-09 16:43:07

It was a little easier after I got 24 zeros for 100! Thereafter, one additional zero each for 125, 250, 375, 500, 625, 750, 875 and 1000 have to be provided. 249 should be the right answer! I didn't know the solution when I posted that question.

Jai Ganesh · 2005-08-09 16:45:18

Problems #n+6
There are 5 boys and 5 girls and they are made to sit in a row such that no two boys and no two girls sit next to each other. How many arrangements are possible?

mathsyperson · 2005-08-09 20:46:36

I get

.

Jai Ganesh · 2005-08-09 21:18:52

And you got it RIGHT!
The Girls can take the odd places and the boys can take the even places.
This is possible in 5! x 5! different ways.
Then, the Boys can take the odd places and the girls can take the even places, which is possible in 5! x 5! different ways.
Hence, the number of arrangements possible is
2 x 5! x 5! = 2 x 120 x 120 = 28,800

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