Maths riddles (of the non-silly variety!)

Eliot Rosewater

I thought it'd be cool if we challenge could each other with semi-serious maths riddles. By semi-serious, I mean not riddles that are merely dainty but ones that involve a bit of real maths, but not riddles that are difficult in the extreme.

Anyhow here's one I made up!

Consider all natural numbers a whose final six digits are 646484. For example:

a=12678390646484
a=2356873213462189465498646484
etc.

Can you find an a such that the cubed root of a is also an integer?

im invisible

two ducks infront of a duck, two ducks behind a duck, and a duck in the middle,
how many ducks?

red_fox

Eliot Rosewater wrote: »

Consider all natural numbers a whose final six digits are 646484. For example:

a=12678390646484
a=2356873213462189465498646484
etc.

Can you find an a such that the cubed root of a is also an integer?

No...?

im invisible

how about this one,
we all know that the square on the hypotenuse is equal to the sum of the squares on the other two sides, and we all know a few integer solutions to this
(x squared + y squared = z squared) eg. 3^2 + 4^2 = 5^2
now if we raise the power, are there any whole number solutions to the equation x^3 + y^3 = z^3
or what about higher powers?

are there any whole number solutions to the equation x^n + y^n = z^n, n>2?

LeixlipRed

im invisible wrote: »

how about this one,
we all know that the square on the hypotenuse is equal to the sum of the squares on the other two sides, and we all know a few integer solutions to this
(x squared + y squared = z squared) eg. 3^2 + 4^2 = 5^2
now if we raise the power, are there any whole number solutions to the equation x^3 + y^3 = z^3
or what about higher powers?

are there any whole number solutions to the equation x^n + y^n = z^n, n>2?

I have a proof but it won't fit in the margin here *dies*

MoogPoo

Eliot Rosewater wrote: »

I thought it'd be cool if we challenge could each other with semi-serious maths riddles. By semi-serious, I mean not riddles that are merely dainty but ones that involve a bit of real maths, but not riddles that are difficult in the extreme.

Anyhow here's one I made up!

Consider all natural numbers a whose final six digits are 646484. For example:

a=12678390646484
a=2356873213462189465498646484
etc.

Can you find an a such that the cubed root of a is also an integer?

I think I have it but its not a very nice proof. I didnt try this too much but was playing with calulator for a while, but i think you cant construct the number. if you had a b = a^3. then the last digit of b must be 4 to get a cube that ends in 4. then you can cube 14, 24, 34, 44.. etc to find cubes that end in 84. this works for 94, or 44. But to get a cube that ends in 484 you could cube 144, 244, 344.... 194, 294, 394... but non of these will end in 484. So no number b can be constructed. I think thats right but i might have missed something.

Would like to see a nice proof thats not trial and error though

MoogPoo

LeixlipRed wrote: »

I have a proof but it won't fit in the margin here *dies*

Dammit, beat me to it

MoogPoo

i thought of this one before: Imagine your in a long hallway with 1000 doors, and your standing at the 100th door we'll say. Your looking for gold or something on one of the 1000 doors but you dont know which one its in. How would you search all the hallway for the gold to get the shortest expectation time? As in would you go backwards checking the 100 doors and then walk back up and check the other 900 or check the 900 first?

Eliot Rosewater

Okay, well I thought the idea would be that someone would post a riddle, and the next one would only be posted when that one was solved, but anyway...

MoogPoo wrote: »

I think I have it but its not a very nice proof.

You're on the right track.

MoogPoo

is it not right? if a = b^3 = xxxx6464684. Then b must end in either xxx44 or xxx94. but none of 144, 244, 344, 444, 544, 644, 744, 844, 944, 194, 294, 394, 494, 594, 694, 794, 894, 994 cubed end in 648. So be cannot be an integer?

Eliot Rosewater

Okay, perhaps you have! It's just hard to see through all the cases!

I'll give the actual proof. We use the fact that if the cubed root of a is an integer then a is a perfect cube.

Now, 2 is a factor of 4 (the last digit of a) so 2 is a factor of a.

4 is a factor of 84 (the last two digits of a) so 4 is a factor of a.

But 8 is not factor of 484 (the last three digits of a) so 8 is a not factor of a.

So [LATEX]\displaystyle a=2^{2}(b)[/LATEX] where b is an odd number. If a was a perfect cube it would have to have a multiple of 3 of every factor (i.e. p.p.p = p^3). But it has only two 2s as factors.

Thus it is impossible to have a as a perfect cube.

MoogPoo

ah yeah thats a lot nicer.

MathsManiac

Given two circles of radius 1, how far apart should their centres be in order that the area of the intersection of the discs is equal to the area of each of the non-intersecting pieces, (i.e., each circle cuts the area of the other in half)?

rjt

MathsManiac wrote: »

Given two circles of radius 1, how far apart should their centres be in order that the area of the intersection of the discs is equal to the area of each of the non-intersecting pieces, (i.e., each circle cuts the area of the other in half)?

Hm. So my attempt (apologies for lack of LaTeX, what's the tag again?):

If we join the two points where the circle meets, we divide the overlap into two identical sectors. The two of these together are half of the area of a full circle (if I read the question right), so each sector is Pi/4 in area. Letting \theta be the angle that subtends the chord

\pi / 4 = 1/2 ( \theta - \Sin(\theta))

So \Sin(\theta) = \theta - \pi / 2. Not sure how to solve this explicitly. Cos(x) = x for about x = .739, so \theta is \pi / 2 + .739. Once you've got this we get cos(\theta / 2) to be 0.40397. So the distance is twice this, .8079

MathsManiac

rjt wrote: »

If we join the two points where the circle meets, we divide the overlap into two identical sectors.

Segments rather than sectors, but I know what you mean!

rjt wrote: »

Not sure how to solve this explicitly.

I think that's because it can only be solved by numerical methods.

Your answer is correct.

By the way, this is one of two versions of Mrs Miniver's problem. The other version asks for the overlap area to equal the sum of the two outer pieces, (i.e., the overlap to be two thirds of one circle rather than half).

Fremen

Prove that at any party, there are at least two people with the same number of friends there.

(Rules: there is more than one person at the party. If person A is friends with person B, then person B is also friends with person A.)

MoogPoo

If theres n people. then each person can have between 0 and n-1 friends. So if the question was false, then 1 person would have 0 friends, 1 person would have 1. etc. But if 1 person has 0 friends. Then the most friends anyone could have would be n-2, because that person is not friends with the 0 friends guy or himself. So we have n-1 people remaining and each can have from 1 to n-2 number of friends. By pigeonhole principle 2 must have the same number of friends.

Fremen

I cut off the top left and bottom right squares on a chessboard, leaving 62 squares. If a domino covers exactly two squares, show that it is impossible to cover this board using 31 dominos.

MoogPoo

a domino will lie on a square and a square adjacent to it. All adjacent squares will be of different colour. But the two squares cut out were both the same colour as they lie on the same diagonal. Therefore the remaining 62 squares have 30black, 32white or other way around. But if you could tile the board you would require 31 of each as the dominos will each tile one of each.

MoogPoo

There are 1000 doors in a hallway and they are initially all closed. You walk up the hallway and open every second door. Then your walk along the hallway again this time for each third door you open it if its closed or close it if its open. Repeat this process for every 4th door, 5th, 6th ....1000th. What doors are left open in the end.

Eliot Rosewater

MoogPoo wrote: »

There are 1000 doors in a hallway and they are initially all closed. You walk up the hallway and open every second door. Then your walk along the hallway again this time for each third door you open it if its closed or close it if its open. Repeat this process for every 4th door, 5th, 6th ....1000th. What doors are left open in the end.

The question's just a little ambiguous. After you walk up opening every second door do you go back to the start of the hall again? Or do you turn around?

If you start again at position one, the answer relies on factors.

Consider door x=a.b.c

It will be opened/closed on the following runs:
a - o
b - c
c - o
ab - c
ac - o
bc - c
abc - o
1 - c
So 8 runs: it will be closed at the end.

It relies on binomial coefficients. Given that the sum of binomial coefficients is even, all doors will be closed at the end.

Except for doors with repeating factors (and door 1). Hmm...

MoogPoo

yeah sorry you start again.
so lets say C is closed and O is open, then its
CCCCCCCCCCCCCCCCCCC.......start
COCOCOCOCOCOCOCOCO.......second
COOOCCCOOOCCCOOOCC........third

your very close with the factors.

Fremen

The doors numbered with a square are left open

I have a biased coin, which lands on heads with an unknown probability (but does produce tails at least sometimes). How do I use this coin to generate a fair coinflip?

rjt

Fremen wrote: »

I have a biased coin, which lands on heads with an unknown probability (but does produce tails at least sometimes). How do I use this coin to generate a fair coinflip?

Flip the coin twice. If it is (heads,tails) call this outcome H, if it is (tails,heads) call this outcome T. Otherwise, repeat the process.

A devious problem - there are 1981 people in a large hall, and each person knows at least 45 other people. (We assume that if A knows B then B also knows A). Show that it is possible to find four people, such that we can seat these four people on a square table such that each person at the table knows the person to their left and right (but not necessarily the person across from them).

MoogPoo

I was going completely the wrong way with the coin one.

I think this is right though.

If A knows B knows D knows C knows A. Then B and C both know A. So then we have to find that for some B and C, they both know another person in common besides A.

So B and C both know a remaining 44 people each. Suppose none of them know someone in common, then there would be 44*44 people required. but 44*44 = 1936 at least needed.

But then there would be only 1981-1936=45 people remaining including D. This means D can only know 44 other people that are not common to B and C. Therefore in order to have any 4 people such they cant sit at the table, you would need 1936 + 46 = 1982 > 1981

rjt

MoogPoo wrote: »

So B and C both know a remaining 44 people each. Suppose none of them know someone in common, then there would be 44*44 people required. but 44*44 = 1936 at least needed.

Hm...

At this stage are B and C fixed? If they are, it's 44+44 people. If not I think it should be 45*44 (45 people that A knows, each of whom knows 44 other different people) = 1980. I don't follow the bit about D then.

The conclusion is a bit suspect anyway - there definitely can be 4 people who can't sit at such a table with 1981 people in the room.

ray giraffe

Fremen wrote: »

The doors numbered with a square are left open

You mean those are not left open!

ray giraffe

rjt wrote: »

A devious problem - there are 1981 people in a large hall, and each person knows at least 45 other people. (We assume that if A knows B then B also knows A). Show that it is possible to find four people, such that we can seat these four people on a square table such that each person at the table knows the person to their left and right (but not necessarily the person across from them).

I think I got it. It's hard to explain without pictures and some definitions but I'll give it a go.

Suppose no such foursome exists.

I am one of those in the hall. I have 45 friends (easier to prove if I have more than 45). How many people are friends of my friends? Each friend lists 44 friends as well as me (Easier if any have extra friends).

Combining the lists that's 45*44= 1980 people, plus myself, 1981 people. Could we have counted any of those 1981 twice? No, because if someone is a friend of 2 of my friends we can go at the square table.

So every one of the 1981 people in the hall is a friend of one of my friends. So each of my friends is a friend of one of my friends. Could one of my friends be a friend of 2 of my friends? No, the four of us would go at the table.

So each of my friends is a friend of exactly one of my friends. Therefore my set of friends is a collection of distinct "friend pairs", so I have an even number of friends. But I had 45 friends! Contradiction.

rjt

ray giraffe wrote: »

I think I got it. It's hard to explain without pictures and some definitions but I'll give it a go.

Suppose no such foursome exists.

I am one of those in the hall. I have 45 friends (easier to prove if I have more than 45). How many people are friends of my friends? Each friend lists 44 friends as well as me (Easier if any have extra friends).

Combining the lists that's 45*44= 1980 people, plus myself, 1981 people. Could we have counted any of those 1981 twice? No, because if someone is a friend of 2 of my friends we can go at the square table.

So every one of the 1981 people in the hall is a friend of one of my friends. So each of my friends is a friend of one of my friends. Could one of my friends be a friend of 2 of my friends? No, the four of us would go at the table.

So each of my friends is a friend of exactly one of my friends. Therefore my set of friends is a collection of distinct "friend pairs", so I have an even number of friends. But I had 45 friends! Contradiction.

Nice. But what if you have more than 45 friends (the question says at least 45)?

Fremen

ray giraffe wrote: »

You mean those are not left open!

Oops, I was counting a #1 run where every door divisible by 1 was opened.

ray giraffe

rjt wrote: »

Nice. But what if you have more than 45 friends (the question says at least 45)?

Suppose I have more than 45 friends. How many people are friends of my friends? Each friend produces a list of at least 44 friends as well as me.

Concatenating the lists that's at least 46*44= 2024 entries. Since there are only 1981 people, we must have a duplicate entry, i.e. someone is a friend of 2 of my friends and can go at the square table.

Here's a nice puzzle

Imagine a 2x1x1 solid cuboid. A spider is on one corner. The spider is clever so he knows the shortest route to take to any point on the box. Which point will take him longest to crawl to?

MoogPoo

nice solution to the last one btw, I really screwed that one up...

Dunno if im missing something. Isnt it the opposite corner?

Fremen

MoogPoo wrote: »

nice solution to the last one btw, I really screwed that one up...

Dunno if im missing something. Isnt it the opposite corner?

No, he can reach that in a distance of 2 root 2 by going diagonally across the faces of two squares. I think the hardest point to reach should be the midpoint of the edge farthest from him, which is a distance root 9.25 from him.

Crap, can't think of any nice problems right now.

MoogPoo

wow i really didnt expect that. Found a cool applet for distances on cuboid if you want to cheat here http://www.btinternet.com/~se16/js/cuboid.htm

resistantdoor

Hey guys, I know this is off topic (and I dont want to hijack the thread) but I have recently been considering doing maths science in university (UCC) and have been looking into whether or not I would be capable of doing well in this course.

My question is: considering I have been looking through this thread and can't understand about 90% of the puzzles and about half of the solutions, would this indicate to you that I would not be capable of such a course.

P.S. I have a strong interest in studying maths and I am a high B/ A in Leaving Cert maths.

Thanks for any help guys.

Fremen

No, these puzzles are designed to be easy to understand but hard to solve, or at least counterintuitive. They're definitely not representative of what you would do in college.

If you can get an A in leaving cert. maths and you're prepared to work hard, you'll probably do ok.

Have a look through the calculus, differential equations and linear algebra courses on khanacademy.org. That's what you'll cover in a first year college course.

Thomas_S_Hunterson

How many points on the planet earth (assuming it is a perfect sphere) have the property that you can travel 1 mile south, 1 mile east and then 1 mile north and be at the same point from which you started?

MoogPoo

1, just the north pole. Wouldnt have got it if you didnt say it was a sphere though.

A king has three advisors. He wants to find out which of them is the smartest so he devises a test. He tells them that he has 3 white hats and 2 black hats. After seating the three facing each other, he tells them that he would randomly pick 3 hats, which would be placed on their heads so that they could each see only the other two hats, but not their own. Unknown to the advisors, the king tosses away the two black hats and only uses the three white hats. After staring at each other for a while, one of the advisors figures it out and shouts, "I know! My hat is white!" How did the advisor know?

Thomas_S_Hunterson

MoogPoo, that is one point but there are others.

Fremen

Thomas_S_Hunterson wrote: »

MoogPoo, that is one point but there are others.

Are there an infinite number? There should be a set of points near the south pole so that I can walk east in a "circle" around the pole in one mile. If I start a mile north of any of those points, I'll be back where I started

Thomas_S_Hunterson

Fremen wrote: »

Are there an infinite number? There should be a set of points near the south pole so that I can walk east in a "circle" around the pole in one mile. If I start a mile north of any of those points, I'll be back where I started

Yup

You could pick any point 1 mile north of a cross section 'near the south pole' which has circumference 1/n for natural n.

equivariant

This is a tricky one - at least it stumped me for a long time

There are 100 people on an island (you are not one of them). Each of them is either a knight or a spy. A knight always answers a question truthfully, whereas a spy may or may not tell the truth (and may answer truthfully sometimes and falsely on other occasions). You know that there are more knights than spies on the island. You are allowed to ask any person on the island whether or not one of the other people on the island is a knight or a spy. You are allowed to do this at most 150 times. Is it possible to determine the true nature of everyone on the island?

equivariant

Fremen wrote: »

No, he can reach that in a distance of 2 root 2 by going diagonally across the faces of two squares. I think the hardest point to reach should be the midpoint of the edge farthest from him, which is a distance root 9.25 from him.

Crap, can't think of any nice problems right now.

Don't think that its the point that you mention Fremen. No point that is on any edge of the block is more than [latex]\sqrt{8}[/latex] away from the ant

I think that the furthest point is a point that is one quarter the way along the diagonal of the square face opposite the start point. In co-ordinates, suppose that the vertices of the block are (0,0,0), (1,0,0), (0,1,0), (1,1,0), (0,0,2), (1,0,2), (0,1,1), and (1,1,2), and that the ant starts at (0,0,0), then the point that is furthest for the ant is (0.75,0.75,2)

BTW - that is a great problem.

RHRN

MoogPoo wrote: »

A king has three advisors. He wants to find out which of them is the smartest so he devises a test. He tells them that he has 3 white hats and 2 black hats. After seating the three facing each other, he tells them that he would randomly pick 3 hats, which would be placed on their heads so that they could each see only the other two hats, but not their own. Unknown to the advisors, the king tosses away the two black hats and only uses the three white hats. After staring at each other for a while, one of the advisors figures it out and shouts, "I know! My hat is white!" How did the advisor know?

(Sorry if my thoughts are disjointed at parts)
Well first, let's say the winning advisor actually has a black hat on. Then, the other two would be able to see a white, and a black hat. But now consider if one of these other two had a black hat on. Then the third advisor could immediately say he had a white. But, neither of them does declare that they have a white.

The winning adivsor thinks to himself "If I had a black hat on, the other advisors should reason, that because neither of them can see two black hats and declare they have a white, only seeing one black, each of them should then reason, that as the other can't defnitively say, it is because they (the advisor the winner is saying is reasoning is) have a white hat on. They should then declare they have a white hat. But they aren't doing this, therefore I must also have a white hat."

I think thats it anyway...

red_fox

My attempt at clarity in the same argument (nothing new in terms of method):

At t=0 Each sees two white hats, so each knows that nobody can see two black hats, so nobody can say what colour they have.

At t=1 Each thinks, if I had a black hat, the other two would see a white and a black, and because nobody spoke up at t=0 that means that they would know that they could not have a black hat themselves so they must have white and will speak up.

At t=2 Each thinks, well since the other two stayed quiet at t=1 that means that I don't have a black hat, so I have a white hat! And so all three declare this at once!