Challange Maths Problems

[Ekin]

I was looking thru a book when i came across these problems. I managed to work most of them out. I'll put some here to see if anyone has lots of free time and do them to keep them occupied.

1. You are given an unlimited supply of 1 x 1, 2 x 2, 3 x 3, 4 x 4, 5 x 5 and 6 x 6 squares.

Two sets of squares are considered the same if the squares can be paired, one from each set, so that the squares in each pair are the same size. For example {1 x 1, 2 x 2, 4 x 4} is the same as {2 x 2, 1 x 1, 4 x 4} but not the same as
{1 x 1, 1 x 1, 2 x 2, 4 x 4}

a) Make a list of all the different sets of 10 squares whose areas add to 48; justify that your list contains all the different sets.

b) Make a list of all the different sets of 10 squares that can be fitted together to cover (without overlaps) a 6 x 8 rectangle; justify that your list contains all the different sets.


2. When Indiana walks into a cave, the door shuts behind him. He looks around and sees chests full of diamonds, emeralds, rubies and sapphires. There is a set of scales near the chests. The following statements are written on the door:

· Each jewel of the same type weighs the same.
· No two different types of jewel weigh the same.
· One sapphire weighs the same as one diamond and one emerald together.
· Three diamonds weigh the same as one sapphire and one ruby.
· One sapphire weighs more than one emerald and three rubies.
· Fourteen rubies weigh more than one sapphire.
· The door will open and all the jewels will be yours if at your first attempt you place one diamond on one tray and other types of jewel on the other tray to make the scales balance.

Of course, Indiana was able to get out of the cave with all the jewels.

a) List the four types of jewel in order from heaviest to lightest. Explain why there is only one answer.

b) If a diamond was placed on a tray scale, what did Indiana place on the other to open the door? Show that this is the only way he could have escapes.


3. Fourteen pebbles is a game played with fourteen pebbles in a jar. To play the game you first roll two dice and add the numbers showing. You remove that number of pebbles. You then choose to roll either one or two dice. If you roll a total equal to the number of pebbles remaining in the jar, you win. Your score is the total you rolled with the second roll multiplied by the number of dice you rolled. If you lose, you score zero.

For example, Bam Bam rolls a 9 with his first roll of two dice. He removes nine pebbles. If he then chooses to roll one die and rolls a 5, he scores 5 x 1 = 5 points.
If he rolls two dice and gets a total of 5, he scores 5 x 2 = 10 points. If he does not roll a total of 5, he loses and scores 0.
a) What non-zero scores are possible in a game of Fourteen Pebbles?

b) Fred and Wilma are playing a game of Fourteen Pebbles and discuss what to do if the first roll is an 11. Fred says “I think you should always choose to roll one die if you have removed eleven pebbles.” Wilma says “No, I think it is better to roll two dice.”

i) If you want to win, that is, to successfully remove the remaining pebbles, as often as possible, who is correct? Explain your answer.
ii) If you want to gain as many points as possible over a long period of time, who is correct? Explain your answer.

c) To obtain the maximum possible score over a long period of time, for what number(s) of pebbles remaining is it better to choose one die for the second roll? For what number(s) of pebbles remaining is it better to choose two dice for the second roll?

[Inn3rchild]

clever way of making people do ur homework

[hobobaggins]

nah. that is his extra credit...

Err... this reminds me of these.. .. .. logic problems.

a) Make a list of all the different sets of 10 squares whose areas add to 48; justify that your list contains all the different sets.

dont you need to just find one, and add another [1x1] to it? or do you add them together?

[Thany]

Originally Posted by Inn3rchild

clever way of making people do ur homework

This does not looks like an homework to me... But still homework can differ when the country change, ne?

[Roots]

Its summer vacation, I refuse to do anything that requires me to think

[Yamano667]

we shouldnt make his homework ..DO IT yourself

[Rhia]

Originally Posted by Inn3rchild

clever way of making people do ur homework

I was gonna say the smae thing...its a very good idea but some people might post the wrong answers on purpose..

[Thany]

Originally Posted by Rhia

I was gonna say the smae thing...its a very good idea but some people might post the wrong answers on purpose..

It's not like there are only mean people, most of them are gone anyway

[mantidor]

this reminds me the times when I went to the physics and mathematics olympiads here, great times, I almost get to travel to Mexico for the finals, but I didn't make the cut the problems you showed are a piece of cake compared with the ones I had to do back in those days *sigh* is so good to remember If I find some of my older problems I'll post them.

[Green²]

(1 2² 3² 4² 5² 6²)

That should make the problem one sets easier to read. Good stuff to know for construction & home remodeling work. Just remember that 5² is the same as twenty five 1x1's.

You're lucky though that they don't make you convert actual area dimensions to nominal dimensions. And most people wonder why they didn't acquire enough material.

[Jinto]

for the problem 1a) I would let your PC do the work use this c-code, compile, run and get the answer:


#include "stdafx.h"
#include "stdio.h"

int x[6];

void square(int i, int y, int count, int w[11])
{
int s;
if(((x[i]+y) < 48) && (count < 10))
{
w[count] = i;
y=0;
for(s=0;s<=count;s++)
{
y += x[(w[s])];
}
count++;
square(i,y,count,w);
}else if(((x[i]+y) >= 48) && (count < 10))
{
if(((x[i]+y) == 48) && (count == 9))
{
printf("%i,%i,%i,%i,%i,%i,%i,%i,%i,%i\n",x[(w[0])],x[(w[1])],x[(w[2])],x[(w[3])],x[(w[4])],x[(w[5])],x[(w[6])],x[(w[7])],x[(w[8])],x[i]);
count++;
square(i,x[i]+y,count,w);
}else if(i < 5)
{
i++;
square(i,y,count,w);
}else if(count > 0)
{
count--;
while((w[count] >= 5) && (count > 0))
{
w[count]=NULL;
count--;
}
if(count >= 0)
{
i = w[count] + 1;
w[count] = i;
y=0;
for(s=0;s<=count;s++)
{
y += x[(w[s])];
}
count++;
square(i,y,count,w);
}
}
}else if(count >= 10)
{
int tmp = i;
count--;
while((count > 0) && (i >= tmp))
{
w[count] = NULL;
count--;
i = w[count];
}
if((count == 0) && (i >= 3))
{
printf("end\n");
}else
{
y=0;
i++;
w[count] = i;
for(s=0;s<=count;s++)
{
y += x[(w[s])];
}
count++;
square(i,y,count,w);
}
}
}

int main(int argc, char* argv[])
{
x[0] = 36;
x[1] = 25;
x[2] = 16;
x[3] = 9;
x[4] = 4;
x[5] = 1;
int a[11];
printf("Starting calculation...\n");
square(0,0,0,a);
return 0;
}


btw. the result will look like:


Starting calculation...
36,4,1,1,1,1,1,1,1,1
25,9,4,4,1,1,1,1,1,1
16,16,9,1,1,1,1,1,1,1
16,9,4,4,4,4,4,1,1,1
9,9,9,9,4,4,1,1,1,1
end


Well that would be my idea to solve this.

[Roots]

Originally Posted by Jinto Lin

for the problem 1a) I would let your PC do the work use this c-code, compile, run and get the answer:

Did I ever tell you that you are my hero Jinto?

[Ekin]

nope, it's not homework. I'm kinda like a maths maniac and was curious cause i couldn't figure them out.

Yea, call me a Maths freak if you want

[_Sin_]

Answer to 2b) He puts 1/3 sapphire and 1/3 ruby on the other tray and laughs at the instruction because the don't say that you have to use intact gems (Take it from me: mathematicians are the laziest kind of human ever )

The reason that there is only one possibility can be deduced from the question itself. If there was more than one possible answer, the question would have been worded differently..^^

[Jinto]

I'ld say he should use 4 rubies.
Why? Lets say diamond = D, emerald = E, rubie = R, sapphire = S

D != E != R != S
all paired combinations are different from each other...
S = D + E
3D = S + R
S > E + 3R
14R > S

E = S-D
=> S > S - D + 3R
=> 0 > -D + 3R
=> D > 3R

S = 3D - R
14R > S
=> 14 R > 3D - R
=> 15 R > 3D
=> 5R > D

=> 3R < D < 5R
lets say D is equal to 4R

Test:

3D = S + R
=> 3*4R-R = S = 11R

14 R > S (yes 14 >11)

S = D + E
11R = 4R + E
=> E = 7R

11R > E + 3R (yes 11>10)

you can check the other conditions too, but I am almost sure this is what was thought to be the solution.

[Robbom]

Does anybody know the answer to three?

[Robbom]

What book was this in?

[Robbom]

Also what are the answers to this?