Puzzles with numbers and things

[snookersfun]

you are totally right, should add 'the even numbers in this problem' (as 2 can't be the sum and I think it is shown for up to quite high even numbers.) Should one never have heard of the fact though, one simply had to list all possible additions and find sums of primes that way.

Anything else sounding strange?


and arghhhh

Almost! 52 (!)

there goes the good impression....lol
I go correcting now...

[davis_greatest]

Nothing else sounded strange.

[davis_greatest]

Logic was part of the math curriculum at my first year in University. It was called 'discreet mathematics.' But in reality it was Basic Logic. (I did fail that course twice.)

Discrete mathematics, I think. It shouldn't be confused with discreet mathematics, which is maths you do without telling anyone about it.

[davis_greatest]

I am deducting myself the 2 points I awarded myself for round 25½ for correctly listing the countries I had been to on holiday.

As, at that stage, I had been the only one posing the questions, I thought it fair to give myself a chance for some points to get on the scoreboard. Now I have got some points elsewhere, I don't think it's right that I should keep those 2 points for answering my own question.

SO HERE IS THE REVISED SCOREBOARD AFTER ROUND 36, with me slipping back down 2 places:

snookersfun……………………….…..16
Vidas……………………………………….8½
abextra……………………………..…...5½
robert602…………………………………5
davis_greatest…………………..……4
elvaago...............................1

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[rambon]

Here's a quick question for you.

I have twelve red balls in my bag. I know, because someone told me, that one of the twelve is a different weight to the other eleven. I don't know whether it is heavier or lighter though.

My question is, using the twelve balls and a set of balancing scales, what is the least number of times I would have to use the scales to determine which ball is the odd one out? Please explain your answer

[davis_greatest]

Here's a quick question for you.

I have twelve red balls in my bag. I know, because someone told me, that one of the twelve is a different weight to the other eleven. I don't know whether it is heavier or lighter though.

My question is, using the twelve balls and a set of balancing scales, what is the least number of times I would have to use the scales to determine which ball is the odd one out? Please explain your answer

I like this question. Remember doing at school, many moons ago. I'll let someone else submit an answer though.

Notes:

1) It can be done, with the same number of weighings, even if you don't know whether the balls are all the same weight or whether there is one that may be lighter or heavier.

2) It can also be done with 12 green balls.

[rambon]

Please note that using twelve green balls means buying 12 sets of balls and is therefore the millionaire's method.....

[davis_greatest]

I have PM'ed you the number of weighings. How you do them is a little longer to write out - there is more than one way.

[elvaago]

Discrete mathematics, I think. It shouldn't be confused with discreet mathematics, which is maths you do without telling anyone about it.

This is actually the first time I tried to write the name of this course in English. In Dutch it's Discrete Wiskunde.

[snookersfun]

Here's a quick question for you.

I have twelve red balls in my bag. I know, because someone told me, that one of the twelve is a different weight to the other eleven. I don't know whether it is heavier or lighter though.

My question is, using the twelve balls and a set of balancing scales, what is the least number of times I would have to use the scales to determine which ball is the odd one out? Please explain your answer

I know that one as well, so I'll pass (although, I would need to rethink, doh)

[davis_greatest]

Here is my answer, in hidden text, which appears when selected. I think I've got this to work! If abextra, robert, or anyone else gets it, they should be given the point too.

(
It can be done with 3 weighings.

Call the balls A,B,C,D,E,F,G,H,I,J,K,L

Step 1 Weigh ABCD v EFGH

Balance -> go to Step 2.1
Don't balance -> call the balls on the heavier side 1,2,3,4 and the balls on the lighter side 5,6,7,8 and go to Step 2.2

Step 2.1 Weigh IJK v ABC
Balance -> go to Step 3.1
IJK heavier -> go to Step 3.2
IJK lighter -> go to Step 3.3

Step 2.2

Weigh 125 v 346
Balance -> go to Step 3.4
125 heavier -> go to Step 3.5
125 lighter -> go to Step 3.6

Step 3.1 Weigh L v A
Balance -> all balls weigh the same.
L heavier -> L is heavier than all the other balls
L lighter -> L is lighter than all the other balls

Step 3.2 Weigh I v J
Balance -> K is the ball heavier than all the other balls
Don't balance -> whichever is the heavier of I and J is the ball heavier than all the other balls

Step 3.3 Weigh I v J
Balance -> K is the ball lighter than all the other balls
Don't balance -> whichever is the lighter of I and J is the ball lighter than all the other balls

Step 3.4 Weigh 7 v 8
whichever is the lighter is the ball lighter than all the other balls

Step 3.5 Weigh 1 v 2
Balance -> 6 is the ball lighter than all the other balls
Don't balance -> whichever is the heavier of 1 and 2 is the ball heavier than all the other balls

Step 3.6 Weigh 3 v 4
Balance -> 5 is the ball lighter than all the other balls
Don't balance -> whichever is the heavier of 3 and 4 is the ball heavier than all the other balls

)

[snookersfun]

I put up a little bit different solution hidden as well. Took me a while, once I got to it though- I didn't remember it to be that twisted.

(
3 weighings are enough.

devide balls into 3 groups
1) weigh 1,2,3,4 vs 5,6,7,8
2) if even, weigh 9,10,11 vs any three above (e.g.1,2,3); if even 12 will be the odd ball
3) if uneven weigh 9 vs 10 (as relative weight will be known by 2), thus either 9,10 or 11


2a) if unbalanced, put l,l,l,h,h vs 9,10,11,12,l (with l-light, h-heavy)
3a) if even, weigh the remaining two heavy balls against eachother or one against any 'norm' ball.
3b) if left side light l1,l2,l3 in question (therefore weigh any two against eachother), if left side heavy h5,h6,l4 questionable (here you have to weigh h5 vs h6)

is that good or not?

)

[davis_greatest]

I put up a little bit different solution hidden as well. Took me a while, once I got to it though- I didn't remember it to be that twisted.

I think I followed it until this bit.... ? What are your h5, h6, l4?

...if left side heavy h5,h6,l4 questionable (here you have to weigh h5 vs h6)

[Semih_Sayginer]

I think I followed it until this bit.... ? What are your h5, h6, l4?

are you playing "battleships"?

[davis_greatest]

Round 38 - The league goes ape

are you playing "battleships"?

That might be worth half a point next time I update the scoreboard...


Here, for a full point, is the next question...


Gordon has just finished organising the Great Ape Snooker League and is working out how much prize money will be needed. He fills lots of sacks with coins, some sacks containing copper coins and some sacks containing gold coins – but each sack containing the same number of coins.

To help him keep track, he writes a day of the week (Monday to Sunday) on each of the sacks with copper coins, each day of the week appearing equally often.

In the league, every ape plays every other ape once in a one-frame match. Each ape gets the same number of sacks of copper coins just for taking part in the competition. The winner of each match gets a sack of gold coins.

Gordon has just finished totting up on his calculator how many coins he will need. "That many?!" he exclaims. "That's terrible!"

"No, it's not, " says Oliver, sitting opposite him, reading the calculator display of the final total: "It's ESIIBL."

How many apes play in the Great Ape Snooker League?