[Guest]

Hey all, you guys are a damn bright bunch, I'm not gonna lie! Those questions were tough buggers and they all got solved so quickly, so well played all!

So here's number 4! Enjoy:

*** When I use root(x), it means the square root of x, root3(x) means cube root of x, rootN(x) is nth root of x etc. ***

1. Suppose a, b, c are the three roots of the equation x^3 - 8x^2 + 5x + 7 = 0. What are the values of:

1a. a + b + c

1b. a^2 + b^2 + c^2

1c. a^3 + b^3 + c^3 ?

2. There are 1988 towns and 4000 roads in a certain country called Amashakashaka! (haha lol) Each road connects two towns, prove that there is a closed path passing through no more than 20 towns.

3. Prove, WITHOUT A CALCULATOR!!! That 7^(root(5)) > 5^(root(7)).

4. The positive integers a and b are such that the numbers 15a + 16 b and 16a - 15b are both squares of positive integers. Find the least possible value that can be taken by the minimum of these two.

(Excited to see how people come up with answers with this one)

5. A problem from 1994, when I was still a wittle 5 year old

1 = 19 - 9 * root(4)

2 = 1 * 9 - 9 + root(4)

3 = -root(1 *　9) + root(9 * 4)

4 = 1^99 * 5

5 = -1 - 9 - 9 + 4!

6 = root(1 + 99) - 4

How much higher can you go? Longest list wins!

See how many you can solve

Weo weo weo, kkkkk GOOOOOOO!!

Edited June 22, 2011 by Twinhelix

[superprismatic]

Before I work on it, I have one question. May the roads cross? If so, may one change the traveled road at a crossroads?

[Guest]

155 = 19 * 9 - (4 squared)

[Guest]

...which I assume to be: given 1988 towns with 4000 roads connecting 2 towns, there must be at least one loop for which I can start traveling in a direction and find myself back where I started within traveling 20 roads. In which case I can not prove it, because I believe it is not true.

First, to understand my approach, I assumed that the more roads you have, the more likelyhood of having a loop smaller than 20 roads. I base this on the the extreme that every town could have a road to every other town. Similar to the handshake problem, there would be n*(n-1)/2 = 1,975,078 roads, and 1988 closed loops of only 2 roads length, let alone the plethora of 3, 4, 5,... & 20 road loops.

So along these lines, I asked myself, "What is the maximum number of roads between 1988 towns where no loop has less than 21 roads. If the answer is 4000 or less, I have the proof. If I find a town & road layout with greater than 4000 roads and no loops less than 21 then I have disproved the statement.

My example (not necessarily the maximum) was to take rows of 11 towns each, each town connected in a line of 10 roads. Each of these rows can then have a road from each end to the same end of every other row such that if I am at town 1 in the row I can travel one road to any other rows first town, travel the 10 roads down to the other that rows town 11, travel one road back to the first row's town 11, then travel the 10 roads back to my starting point for a total trip of 22 roads. This layout requires you to travel through at least two rows of 11 towns no matter what loop you take, so 22 roads is the shortest loop. Now this layout would have 180 rows of 11 towns (with 8 towns remaining that we will mention later). So each row has 10 roads for 1800 roads, plus each row has n(n-1)/2 roads to other rows or 16110 roads, times 2 since a row has roads on each end (kinda like shaking everyones right hand with your right, left hand with your left, and no left-right mixing), for 32220, plus the 8 "loner" towns that each have a road to the rows but not to each other for 8*180=1440. This gives us 1800+32220+1440 = 35,460 roads with no 20 road loops. Now I send in the US Air Force to randomly carpet bomb all but 4000 of them, and the statement is disproved.

...That all depends on Superprismatics crossroad clarification, and the assumption that the name of the country has nothing to do with the problem. But an excellent puzzle regardless.

Edited June 23, 2011 by Copernicus

[Guest]

Before I work on it, I have one question. May the roads cross? If so, may one change the traveled road at a crossroads?

No sorry they do not cross.

[Guest]

...which I assume to be: given 1988 towns with 4000 roads connecting 2 towns, there must be at least one loop for which I can start traveling in a direction and find myself back where I started within traveling 20 roads. In which case I can not prove it, because I believe it is not true.

First, to understand my approach, I assumed that the more roads you have, the more likelyhood of having a loop smaller than 20 roads. I base this on the the extreme that every town could have a road to every other town. Similar to the handshake problem, there would be n*(n-1)/2 = 1,975,078 roads, and 1988 closed loops of only 2 roads length, let alone the plethora of 3, 4, 5,... & 20 road loops.

So along these lines, I asked myself, "What is the maximum number of roads between 1988 towns where no loop has less than 21 roads. If the answer is 4000 or less, I have the proof. If I find a town & road layout with greater than 4000 roads and no loops less than 21 then I have disproved the statement.

My example (not necessarily the maximum) was to take rows of 11 towns each, each town connected in a line of 10 roads. Each of these rows can then have a road from each end to the same end of every other row such that if I am at town 1 in the row I can travel one road to any other rows first town, travel the 10 roads down to the other that rows town 11, travel one road back to the first row's town 11, then travel the 10 roads back to my starting point for a total trip of 22 roads. This layout requires you to travel through at least two rows of 11 towns no matter what loop you take, so 22 roads is the shortest loop. Now this layout would have 180 rows of 11 towns (with 8 towns remaining that we will mention later). So each row has 10 roads for 1800 roads, plus each row has n(n-1)/2 roads to other rows or 16110 roads, times 2 since a row has roads on each end (kinda like shaking everyones right hand with your right, left hand with your left, and no left-right mixing), for 32220, plus the 8 "loner" towns that each have a road to the rows but not to each other for 8*180=1440. This gives us 1800+32220+1440 = 35,460 roads with no 20 road loops. Now I send in the US Air Force to randomly carpet bomb all but 4000 of them, and the statement is disproved.

...That all depends on Superprismatics crossroad clarification, and the assumption that the name of the country has nothing to do with the problem. But an excellent puzzle regardless.

Hey your assumptions are correct. However I think your approach is not the best.

Here's a hint:

See the towns as vertices, the roads as edges. Then try proof by contradiction, assume that all closed loops are 21 or more.

[superprismatic]

...which I assume to be: given 1988 towns with 4000 roads connecting 2 towns, there must be at least one loop for which I can start traveling in a direction and find myself back where I started within traveling 20 roads. In which case I can not prove it, because I believe it is not true.

First, to understand my approach, I assumed that the more roads you have, the more likelyhood of having a loop smaller than 20 roads. I base this on the the extreme that every town could have a road to every other town. Similar to the handshake problem, there would be n*(n-1)/2 = 1,975,078 roads, and 1988 closed loops of only 2 roads length, let alone the plethora of 3, 4, 5,... & 20 road loops.

So along these lines, I asked myself, "What is the maximum number of roads between 1988 towns where no loop has less than 21 roads. If the answer is 4000 or less, I have the proof. If I find a town & road layout with greater than 4000 roads and no loops less than 21 then I have disproved the statement.

My example (not necessarily the maximum) was to take rows of 11 towns each, each town connected in a line of 10 roads. Each of these rows can then have a road from each end to the same end of every other row such that if I am at town 1 in the row I can travel one road to any other rows first town, travel the 10 roads down to the other that rows town 11, travel one road back to the first row's town 11, then travel the 10 roads back to my starting point for a total trip of 22 roads. This layout requires you to travel through at least two rows of 11 towns no matter what loop you take, so 22 roads is the shortest loop. Now this layout would have 180 rows of 11 towns (with 8 towns remaining that we will mention later). So each row has 10 roads for 1800 roads, plus each row has n(n-1)/2 roads to other rows or 16110 roads, times 2 since a row has roads on each end (kinda like shaking everyones right hand with your right, left hand with your left, and no left-right mixing), for 32220, plus the 8 "loner" towns that each have a road to the rows but not to each other for 8*180=1440. This gives us 1800+32220+1440 = 35,460 roads with no 20 road loops. Now I send in the US Air Force to randomly carpet bomb all but 4000 of them, and the statement is disproved.

...That all depends on Superprismatics crossroad clarification, and the assumption that the name of the country has nothing to do with the problem. But an excellent puzzle regardless.

The flaw in your argument is that you can't construct those 16110 roads between lines. In a planar graph, this can't be done without crossings of roads.

[Guest]

The flaw in your argument is that you can't construct those 16110 roads between lines. In a planar graph, this can't be done without crossings of roads.

Okay imagine there's bridges where roads would cross

[dark_magician_92]

see next post

Edited June 25, 2011 by dark_magician_92

[dark_magician_92]

#2

a+b+c = 8, sum of roots tsken 1 at a time

part (ii) = (a+b+c)^2 - 2*(ab+bc+ca)

= 64 - 2*(sum of roots taken 2 at a time = -5)

=74

for (iii) a^3+ b^3 +c^3 -3abc= (a+b+c)(a^2+b^2+c^2 - ab -bc - ca)

a^3+b^3+c^3 = 3(product of all roots = -7) + 8*(74+5)

= -21+632

=611

[Guest]

#2

a+b+c = 8, sum of roots tsken 1 at a time

part (ii) = (a+b+c)^2 - 2*(ab+bc+ca)

= 64 - 2*(sum of roots taken 2 at a time = -5)

=74

for (iii) a^3+ b^3 +c^3 -3abc= (a+b+c)(a^2+b^2+c^2 - ab -bc - ca)

a^3+b^3+c^3 = 3(product of all roots = -7) + 8*(74+5)

= -21+632

=611

Can you show working?

[plainglazed]

if the floor function is allowed: 155 = 1 * root(9)! * floor(root((root(9)!)!)) - floor(root(root(4))

or if multiple factorials are allowed: 155 = 1 * 9!! ! - root(9) - 4

Edited June 29, 2011 by plainglazed

[Guest]

I can not find the complete answer for this problem.

Here just some hint.

The maximum length for the circle looping 1998 cities is 1997 roads.

Of course, this is not the case for the problem context.

Now, think there is two group. one is 21 cities. and other is 1977 cites, both of them is closed. if assume the two groups are isolated. The road length requirement is 20 + 1976 = 1976 roads. However, to keep max loop is 21 cites, we could add more lines between cites between two groups. Label A as 1,2,…1977. B as 1,2,…21.

First connect A1 and B1. Do not break assumption.

2. connect A19 and B1. The loop B1-A1-A19-B1, still 21 road.

3.conect A37 and B1. still right.

4, keep continue until up to 1960. the last one is 1945

Totally 108 roads.

Now how about B2. B2 can not connect to A group, otherwise there is closed circle then than 21 length. B3. same.

However, B4 could be connect to middle of A1945-A1. the circle is 21. so to B7, B10,B13,B16,B19, totally 16 roads.

Besides, A10 can connect to middle point ot A1-19, A19-A37.. totally 107 roas.

In all 108 + 107 _+ 6= 221 roads.

Now we think group A(56), B(21), C(21)

A and B. reduce 3 roads from case 1.

A and C reduce 3 roads from case.

B and C can connect 2 roads.

totally 2* 221 - 3*2 + 2 = 438

….

no idea right now.