Logic/maths puzzles

[Charlie]

Well the vast majority of people in the world have 2 arms.

A few people have less than two arms (amputation and such).

 

Therefore the average number of arms is going to be a little bit less than 2.

The next person you meet is very likely to have 2 arms.

 

 

 

2, 3 and 6?

 

2x3 = 6

6x6 = 36

 

 

 

I think I must be missing something in both of them...

[MoogleViper]

1 is correct. 3 isn't.

[Jonnas]

Most people have 2 arms. Some have 1 or none, therefore the average is somewhere below 2.

 

Therefore, the odds to meet someone with 2 arms is Very Likely.

 

EDIT: And I see I was right with this

 

 

 

Listing several possibilities:

 

Age Combination / Sum

 

433 / 10

623 / 11

922 / 13

661 / 13

941 / 14

12 3 1 / 16

18 2 1 / 21

 

We see that both 9-2-2 and 6-6-1 have the same sum. The census taker sees that the next door has the number "13" on it, and struggles to decide wether it's one or the other.

He goes back to the woman, and she mentions an older child. With this, he knows that the older child can't be the 6 year old twins, so he knows that the older child is 9 years old, and the younger siblings are 2 year old twins.

The combination is 9-2-2

 

Conclusion: Don't screw up with the heads of census takers. Seriously, they're just trying to do their jobs.

 

 

 

As for Problem 2, I'll need to think a bit more.

Edited February 6, 2010 by Jonnas

[chairdriver]

Ok well here's a maths one that I've found.

 

There are one thousand lockers and one thousand students in a school. The head teacher asks the first student to go to every locker and open it. Then he has the second student go to every second locker and close it. The third goes to every third locker and, if it is closed, he opens it, and if it is open, he closes it. The fourth student does this to every fourth locker, and so on. After the process is completed with the thousandth student, how many lockers are open?

 

This exact question was in my Oxford entrance exam!

 

I can't remember the answer, but I remember working with mods alot.

[MoogleViper]

Most people have 2 arms. Some have 1 or none, therefore the average is somewhere below 2.

 

Therefore, the odds to meet someone with 2 arms is Very Likely.

 

EDIT: And I see I was right with this

 

 

 

Listing several possibilities:

 

Age Combination / Sum

 

433 / 10

623 / 11

922 / 13

661 / 13

941 / 14

12 3 1 / 16

18 2 1 / 21

 

We see that both 9-2-2 and 6-6-1 have the same sum. The census taker sees that the next door has the number "13" on it, and struggles to decide wether it's one or the other.

He goes back to the woman, and she mentions an older child. With this, he knows that the older child can't be the 6 year old twins, so he knows that the older child is 9 years old, and the younger siblings are 2 year old twins.

The combination is 9-2-2

 

Conclusion: Don't screw up with the heads of census takers. Seriously, they're just trying to do their jobs.

 

 

 

As for Problem 2, I'll need to think a bit more.

 

That is correct.

 

This exact question was in my Oxford entrance exam!

 

I can't remember the answer, but I remember working with mods alot.

 

Grunch already posted the answer last page if you want to know it.

[Supergrunch]

This exact question was in my Oxford entrance exam!

 

I can't remember the answer, but I remember working with mods alot.

If you just look at even/odd factors, you can do it without modular algebra at all, but I guess it amounts to a similar kind of thing.

Ok some more. (spoilered to look tidier.)

 

What is the probability that the next person you meet has an above average number of arms?

 

Impossible, Very Unlikely, Unlikely, Even, Likely, Very Likely, Certain.

 

 

* There are 10 baskets containing apples.

* There are various amounts of apples in each basket ranging from 10 to 20.

* 9 of the baskets contain apples weighing 4 ounces each.

* 1 of the baskets contains apples weighing 5 ounces each.

* All the apples look the same.

* The equipment you have is a set of scales and an empty basket.

* It is late and the lorry is waiting to take the apples to market. You only have time to make one measurement using the scales.

 

Can we find out which basket has the heavier apples? If so how?

 

 

A census taker approaches a house and asks the woman who answers the door,"How many children do you have, and what are their ages?"

 

Woman: "I have three children, the product of their ages are 36, the sum of their ages are equal to the address of the house next door."

 

The census taker walks next door, comes back and says, "I need more information."

 

The woman replies, "I have to go, my oldest child is sleeping upstairs."

 

Census taker: "Thank you, I now have everything I need."

 

What are the ages of each of the three children?

I didn't actually manage to solve any of these. Not a fan of the first as it basically relies on "average" being interpreted as "mean", which is admittedly the most common usage, but if the question was posed with "mean" instead, the answer would be pretty obvious.

 

The third is cool, although I was thinking about age as a continuous rather than a discrete variable so got myself in a mess - from this perspective even twins differ in age.

 

And for the second I caved and looked up a solution because I was stuck. I was actually thinking the scale was just a two-way balance, so I think it's worth noting that it tells you the actual weight of what you put on it, rather than just giving comparisons, as that's necessary for the solution. Which is awesome, incidentally.

[MoogleViper]

I didn't actually manage to solve any of these. Not a fan of the first as it basically relies on "average" being interpreted as "mean", which is admittedly the most common usage, but if the question was posed with "mean" instead, the answer would be pretty obvious.

 

It's more of a, interesting how it works/not what a lot of people would originally think.

 

And for the second I caved and looked up a solution because I was stuck. I was actually thinking the scale was just a two-way balance, so I think it's worth noting that it tells you the actual weight of what you put on it, rather than just giving comparisons, as that's necessary for the solution. Which is awesome, incidentally.

 

Done it.

[Zell]

I was thinking the same thing Supergrunch, but now you've said that I've found a solution.

 

Label the baskets 1 to 10. Put 1 apple from basket 1 into the empty basket, 2 from basket 2, 3 from basket 3 and so on. We now have 55 apples in the basket. If all the apples weighed 4oz, we would expect a reading of 220oz. However some of the apples weigh 5oz so the reading will be a bit greater. How much greater depends on what basket contains the heavier apples. If the reading is 224, then we deduce that 4 of apples in the basket weigh 5oz and as we took 4 apples from basket 4, we conclude that the basket with the heavier apples is basket 4. In general, the difference between the actual scales reading and 220 equals the number of the basket which contains the heavier apples.

 

It actually doesn't matter what quantity of apples you take from each basket, as long as they're all different.

 

 

A fairly easy one:

 

Two astronauts arrive on an alien planet. The astronauts know there are two races of aliens on this planet, Veraciters and Gibberish. Physically they are indistinguishable, but Veraciters always tell the truth and Gibberish always lie.

 

Three aliens appear. The astornauts give the aliens the names Alfy, Betty and Gemma.

 

The astronauts ask Alfy: "To which race does Betty belong?" Alfy replies: "Gibberish."

The astronauts ask Betty: "Do Alfy and Gemma belong to different races?" Betty replies: "No."

The astronauts ask Gemma: "To which race does Betty belong?" Gemma replies: "Veraciter."

 

Which race does each alien belong to?

[MoogleViper]

Betty must be lying as Alfy and Gemma giver different answers. Therefore she is Gibberish. Which means that Alfy is telling the truth (Veraciter) and Gemma is lying (Gibberish).

 

[chairdriver]

If you just look at even/odd factors, you can do it without modular algebra at all, but I guess it amounts to a similar kind of thing.

 

Reading your answer I do remember getting 31. I think that was one of the few answers I properly got.

[Jonnas]

That really changes everything:

 

Assuming the ten baskets are properly labeled, we take 10 apples from one, 9 apples from the second, 8 from the third, and so on, and we put them all on the empty basket.

We take the measurement of the basket-formerly-known-as-empty, and see the results.

 

Now, if every apple weighed 4 ounces, the weight should be 220 ounces. But since there are a few apples that weigh 1 ounce more each, we'll get a value that's a bit higher.

From said value, we know how many 5 ounce apples there are in the basket. If the weight is 221, there is one 5 ounce apple. If the weight is 230, there are ten 5 ounce apples.

 

Now that I finished typing this, I realise that we could add 0 to 9 apples instead of 1 to 10. It would certainly be more practical.

 

 

A fairly easy one:

 

Two astronauts arrive on an alien planet. The astronauts know there are two races of aliens on this planet, Veraciters and Gibberish. Physically they are indistinguishable, but Veraciters always tell the truth and Gibberish always lie.

 

Three aliens appear. The astornauts give the aliens the names Alfy, Betty and Gemma.

 

The astronauts ask Alfy: "To which race does Betty belong?" Alfy replies: "Gibberish."

The astronauts ask Betty: "Do Alfy and Gemma belong to different races?" Betty replies: "No."

The astronauts ask Gemma: "To which race does Betty belong?" Gemma replies: "Veraciter."

 

Which race does each alien belong to?

 

Alfy and Gemma have conflicting answers, so they're of different races, which means Betty lied. Betty is Gibberish.

Having this in mind, Alfy tells the truth and Gemma lies. Alfy is a Veraciter and Gemma is also Gibberish

 

[Supergrunch]

Betty must be lying as Alfy and Gemma giver different answers. Therefore she is Gibberish. Which means that Alfy is telling the truth (Veraciter) and Gemma is lying (Gibberish).

Alfy and Gemma have conflicting answers, so they're of different races, which means Betty lied. Betty is Gibberish.

Having this in mind, Alfy tells the truth and Gemma lies. Alfy is a Veraciter and Gemma is also Gibberish

Zell hasn't confirmed it, but these answers have got to be right.

 

ANYWAY! New problem time. You have a rectangular chocolate bar with an unknown number of squares, let's say it's m x n (in terms of squares, not other dimensions). To break it up right down into its constituent squares, you'll of course have to make multiple breaks, but the question is: what is the minimum number of breaks required to do this? (in terms of m and n) Note that your answer has to include an explanation of why you can't do it with fewer breaks.

[MoogleViper]

Are we allowed to stack the chocolate as we break it?

[SPAMBOT4000]

Before I embarrass myself by trying to explain an incorrect answer is it m-1 * x-1 ?

[Supergrunch]

Are we allowed to stack the chocolate as we break it?

Nope, only the most basic type of break is allowed.

Before I embarrass myself by trying to explain an incorrect answer is it m-1 * x-1 ?

Well, there's not an x in the equation, but I guess you mean (m-1)(n-1), which isn't right.

 

Edit: Ah, might have confused you. When I said "m x n" I just meant "m by n" - there are m squares one way and n the other.

[SPAMBOT4000]

Yeah not sure why I put an x instead of an n. My only defence is that I was pretty tired when I posted! Anyway imagining breaking various small sizes of chocolate bar in my head it seems to be mn - 1 ?

IF that's right this time I can sorta appreciate why but would probably have trouble explaining it!

[weeyellowbloke]

I think Spambot is nearly right, is it (m-1)+(n-1). For example a rectangular bar consisting of m=6, n=3 would have 5 divisions along m side and 2 divisions along the n side. You could do it with more breaks by snapping the longer dimension first and then breaking each shorter dimension slab individually, which would be (m-1)+(n-1*m) I think (assuming m is the longer dimension).

 

Hmmm, poorly explained and could be completely wrong

Edited February 24, 2010 by weeyellowbloke

[Dannyboy-the-Dane]

I started thinking about this problem, and now I want a bar of chocolate so much ...

[Mr_Odwin]

The answer is mn-1.

 

Say you break the m side first into m strips. This takes m-1 breaks.

Then there are m strips of n length which each take n-1 breaks to get down to constituent parts.

 

So you have m-1 + m(n-1) = mn-1.

[Zell]

Assume m>n. Starting with snapping widthways (ie if you had a 2 x 8 bar you would get two 2 x 4 bars) we have to snap a minimum of m - 1 times. We are left with m bars of dimension 1 x n. We now have to snap each of the remaining bars n - 1 times. So the total number of snaps is (m - 1) + m*(n - 1). If we snapped lengthways first then we would get an answer of (n - 1) + n*(m - 1). Expanding both of these out we find that the solutions are equal (= mn - 1), so it doesn't matter which way we snap first.

 

Regarding my last puzzle, the answers were both correct.

 

EDIT: Beaten to it.

[chairdriver]

I had that answer, and thought there was some trick to it to get less...

[Supergrunch]

You've both shown one (or in Zell's case, two) way(s) of breaking up the bar, but neither of you have proved why the number you get is the minimum number of breaks:

Note that your answer has to include an explanation of why you can't do it with fewer breaks.

[MoogleViper]

It's actually possible to do it in one break.

 

 

[dwarf]

You could have an oblong bar of chocolate with just 2 chunks, therefore you could just split it up into 2 chunks with 1 break. My method is the least amount of breaks because it isn't really possible to get less than 1 break unless you're being silly.

 

Obviously I don't know.

 

However, I do agree with Moogle's method. In any case I'm interested to see the solution, because I am always confused as to the quickest method of cutting a piece of paper into chunks when there's a 2 by x table on it. (x being any number above 2) I swear I always cut take the most amount of cuts to cut the table.