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#201 Re: Puzzles and Games » Ping pong tournament » 2012-06-02 21:33:14
143286759
#202 Re: Puzzles and Games » Ping pong tournament » 2012-06-01 18:33:26
Let's start with
123456789 where 3, 6 and 9 are the arbitrators.
Then 132465798
231564897 where 1 4 and 7 have played 2 games, in order for them to be allowed to arbitrate.
That was the easy part, working on the next steps 
#203 Re: Puzzles and Games » Ping pong tournament » 2012-06-01 18:15:51
Hi Bobbym,
I didn't say it is easy After all, we are not here for the easy ones
This one, however, does have a solution because it was published it a riddles site.
Have a nice weekend!
#204 Re: Puzzles and Games » Ping pong tournament » 2012-06-01 02:24:43
Well, you are right; actually I was considering the first 3 conditions as one These 3 can't be broken.
The 4th condition, which is "after each player arbitrates one game, he must play at least 2 times against another athlete before being allowed to arbitrate again", can be broken, but we request that this happens the least number of times.
#205 Puzzles and Games » Ping pong tournament » 2012-05-31 19:40:19
- anna_gg
- Replies: 10
9 ping pong players will participate in a tournament. There are only 3 tables where 3 games can be played simultaneo
usly. Two players will be playing in each game, while a third will be acting as the arbitrator. For example, the first round would be 12 3 45 6 78 9 with 3, 6 and 9 being the arbitrators and 12 45 78 playing against each other.
There are two rules for the tournament: It must be completed in 12 rounds of 3 simultaneous games, where each player will play against each of the other 8 only once, and will be arbitrating exactly 4 games. Moreover, after each player arbitrates one game, he must play at least 2 times against another athlete before being allowed to arbitrate again.
You will realize that it is impossible to have all two conditions met together. Can you write a schedule that would meet the first condition and would break the second condition for a minimum number of times? The answer must be 12 rows of 9 digits each, where the 3rd, 6th and 9th digit of each row will be the arbitrator, while all the others will be the players playing against each other, e.g. 12 3 45 6 78 9 for the first round (1 is playing against 2 and 3 arbitrates, 4 against 5 etc).
#206 Re: Puzzles and Games » Perfect Squares » 2012-04-16 03:48:38
Here is my solution: It is based to the fact that each perfect square N^2 is the sum of the first N odd numbers (5^2 = 25 = 1+3+5+7+9).
Thus the difference of any 2 perfect squares should equal to the sum of consecutive odd numbers (and this should equal 60).
Starting from 1, we write down the sums of the odd numbers:
1+3+5+7+9+11+13 = 49 while 1+3+5+7+9+11+13+15 = 64. Thus we cannot make 60 starting from 1.
We do the same with 3: 3+5+7+9+11+13=48 while 3+5+7+9+11+13+15 = 63. Not possible.
Starting from 5:
5+7+9+11+13+15=60 Here we are.
So, one perfect square is 4 and the next is 64, their difference being 60, so the first number we are looking for is 34 (34-30 = 4 and 34+30 = 64, both of them perfect squares).
Similarly, we find that the only other sum of consecutive odd numbers equaling 60 is 29+31.
Therefore one perfect square is 1+3+5+...+27=196 (14^2) and the next is 1+3+5+...+27+29+31=256 (16^2) and the second number we are asking for is 226 (226-30 = 196, 226+30 = 256).
There is no other series of successive odd numbers equaling 60, so these two are the only numbers with this property.
Obviously this is not a proper "proof"; it is more based on a "guess and try" method, but it works!
#207 Re: Puzzles and Games » Weighings - A sack of flour » 2012-04-15 11:57:20
Funny thing, I posted this puzzle two months ago and nobody had replied till yesterday! Once I dropped my first line, this triggered 56 posts! Thank you guys!!!
#208 Re: Puzzles and Games » Weighings - A sack of flour » 2012-04-15 11:54:41
I had done the same mistake when I first fell onto this puzzle. I don't think we (=you) are completely the wrong way. I think the only constraint is that we cannot use each bag (weight) more than once in the same weighing.
#209 Re: Puzzles and Games » Weighings - A sack of flour » 2012-04-14 22:15:30
These are "possible" alternative weighings, so why not use them?
To make 9500, for example, you take the total quantity of 10000 (no need to weigh it, as we already know it's 10000) and remove the quantity required to balance the 200 and 300 g weights together; so the remaining is 9500.
Same way with 9700 (we use only the 300g weight), 9800 (only the 200g weight), 9900 (we use both, one in each pan, to weight 100g, so the remaining is 9900).
#210 Re: Puzzles and Games » Weighings - A sack of flour » 2012-04-14 21:42:59
LATER EDIT: I am in the process of correcting this 'solution'. I am optimistic that I'll get there some time (... length of time unspecified ...)
I am assuming the grocer has a limitless supply of weightless bags to put his derived flour weights into.
How about subscripts?
A weight that exists at the start will have zero as its subscript.
A weight derived at 1st weighing will have subscript 1.
I did have some others here, not multiples of 100. I have now removed these.
I do not think something like 4850 can later generate a multiple of 100 except by restoring the number that made it.
This conjecture needs proving (or a counter example).I believe this is a complete list of 1st weighings.
These 1st weighing amounts are put into bags and become additional 'weights'.
Now, what can be made at the 2nd weighing?
Theorem: (i) Where different flour bags are used the subscript total to achieve a new weight must be strictly less than the subscript of the new weight. *
(ii) but if the same bag is used in several stages the subscript total to achieve a new weight need not count this bag repeatedly
to make this total strictly less than the subscript of the new weight. An example of this is shown later in achieving 1400_3 and 1600_3.This theorem needs to be modified. My thanks to anonynmstefy for pointing this out.
* so here, 2 > 1 + 0 + 0
Now those 'weights' can also be bagged up to use at the next stage.
3rd weighings
I cannot find a way to do 1400.
But anoninmstefy did:
I think that's it. If you spot any more, please let me know.
Oh darn it! I've just noticed that 5000. That'll be 5000 against 5000.
So the list needs some additions. ..... Tomorrow perhaps. No better still, you've seen the method, I'll leave it as an exercise for the reader.
Additionally, for whatever amount, x, that can be measured out, the amount 10000 - x is also available for sale which doubles the list!
List so far:
I'm assuming the probability is easy to get from here.
Bob
Why not also
10000_1
5000_1
100_1
9900_1
200_1
9800_1
4900_1
5100_1
300_1
9700_1
500_1
9500_1
#211 Re: Puzzles and Games » Weighings - A sack of flour » 2012-04-14 21:31:27
Hi anna_gg;
You eliminated all the possible weights that are not multiples of 100. Is that okay?
Yes, because the problem says the grocer only sells in multiples of 100 grams.
#212 Re: Puzzles and Games » Weighings - A sack of flour » 2012-04-14 04:46:32
OK. By the 1st weighing we can have:
0g
10000g
5000g
100g
9900g
200g
9800g
4900g
5100g
300g
9700g
500g
9500g
Then we must make combinations of the existing balance weights and the ones we have created.
#213 Puzzles and Games » Perfect Squares » 2012-04-10 20:21:03
- anna_gg
- Replies: 20
Consider an integer x. If we add 30, then the result is a perfect square. If we subtract 30, the result is also a perfect square. How many such integers are there?"
#214 Puzzles and Games » Weighings - A sack of flour » 2012-02-19 05:31:44
- anna_gg
- Replies: 57
A grocer has a sack of flour which weights 10 kgs, 2 weights of 200 and 300 grams and a 2-pan balance.
He only sells the flour at multiples of 100 grams, and only if he can measure the desired quantity in only 3 weighings.
What is the probability that he can serve the first customer that enters the shop, who may desire to buy any quantity, starting from 100 grams and up to 10 kgs?
Suppose that each weighing is completed once the 2 pans balance out and also that the bags have negligible weight.
#215 Re: Puzzles and Games » The Egg-seller's eggs.. » 2012-01-28 11:21:25
Suppose we are looking for x, where x is the number of eggs. Then x+1 should be a multiple of 2,3,4,5,6.
The smaller number to meet this criteria is 5x12 = 60, then 120, then 180 etc.
Moreover, x should be a multiple of 7, thus (x+1)mod7 = 1, thus 120-1 = 119, 238 etc.
The number also has to be <150, so it is 119.
#216 Re: Puzzles and Games » 5 balls with 5 weighings » 2012-01-28 04:42:09
I started by weighing 2 vs 2, but we have 15 different possibilities:
1st plate 2nd plate
10+20=30, 30+40=70
10+20=30, 30+50=80
10+20=30, 40+50=90
10+30=40, 20+40=60
10+30=40, 20+50=70
10+30=40, 40+50=90
10+40=50, 20+30=50
10+40=50, 20+50=70
10+40=50, 30+50=80
20+30=50, 10+50=60
20+30=50, 40+50=90
10+50=60, 20+40=60
10+50=60, 30+40=70
20+40=60, 30+50=80
30+40=70, 20+50=70
In cases 7, 12 and 15 the scale balances,
but then what?
#217 Puzzles and Games » 3 logicians - 8 stamps » 2012-01-22 11:13:57
- anna_gg
- Replies: 6
Here is another:
3 logicians and one moderator have decided to play a game. The moderator has a set of 8 stamps, of which 4 are red and 4 are black. He affixes two stamps to the forehead of each logician, so that each of them can see all the other stamps except those two in the moderator's pocket and the two on his or her own head. He then asks them in turn if they know the colors of their own stamps. Target of the game is to have at least one of the logicians guess the correct colors of his own dots (and of course explain his reasoning). If the 1st player does not know the answer, the 2nd is asked and then the 3rd, then 1st again etc (1-2-3-1-2-3 etc).
Before the beginning of the game, the first logician is asked to which of the 3 positions he wants to sit. What should be his choice and why?
#218 Re: Help Me ! » Make 1000 using 6 8's » 2012-01-13 08:30:09
Thank you all for your help!!
#219 Re: Help Me ! » Make 1000 using 6 8's » 2012-01-12 08:20:55
Confirmed that powers are allowed!
#220 Re: Help Me ! » Make 1000 using 6 8's » 2012-01-12 02:57:03
Not sure about powers; I must confirm this with the friend who gave me the problem.
Either way, I could not go that close myself!
#221 Re: Help Me ! » Make 1000 using 6 8's » 2012-01-12 02:46:00
Awesome! Thanks!!!
#222 Re: Introductions » Happy New Year! » 2012-01-12 01:18:09
@MathIsFun: Thanks for your comforting and inspiring words! Fortunately, I am not in my 90s yet!!
@bobbym: But I have to keep my mind from getting lazy!
#223 Introductions » Happy New Year! » 2012-01-11 22:28:13
- anna_gg
- Replies: 6
In an effort to revive my grey matter and hoping to somehow delay the signs of ageing (and the Alzheimers' disease who is definitely on the way!!!), I have recently joined your team!
I am amazed by the devotion all of you demonstrate! I posted my first 2 questions yesterday, and I already have some of you working on them!
I am 48, hold a polytechnic degree in Naval Engineering and work in IT for many years. Btw, I am Greek
Thanks for accepting me in your nice group! 
Anna
#224 Re: Help Me ! » Make 1000 using 6 8's » 2012-01-11 09:48:40
We are only allowed to use decimals in the form 8.8, not .8, as this assumes leading zeros 
Otherwise, another like that would be 888/.888.
#225 Re: Puzzles and Games » 5 balls with 5 weighings » 2012-01-11 05:14:36
I have gone through a similar problem "Again the puzzle of balls and scales", which is listed here under the same category, but can't follow the same analysis for the 5x5 problem...