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#101 Re: Help Me ! » Ice cream toppings » 2016-01-14 02:34:24
Great job!
#102 Re: Help Me ! » Ice cream toppings » 2016-01-13 03:01:29
@Phrontister: DABC with ABCN have 3 toppings in common, which, unfortunately, is not allowed 
I am stuck in the 13 combinations, although I am sure that the correct answer is 14.
BTW, and sorry to sound like an idiot, what is "M"???
#103 Re: Help Me ! » Ice cream toppings » 2016-01-12 09:02:18
The answer is 14, I have represented each topping with a letter; so we have letters A B C D E F G H I J K L M and N, but I am trying to find the 14 possible combinations (with maximum 1 letter in common) and I still can't!!
#104 Re: Help Me ! » Ice cream toppings » 2016-01-12 08:05:40
The last bit,
also the cones must have at maximum one common topping with each other?
is what makes the answer so short.
What is this license problem?
The license problem is the following (with several variations):
In Riddleland, the license plates are composed of 8 digits, each from 0 to 9. All plates must differ from each other in at least 2 of the 8 positions. What is the maximum number of plates that can be issued?
#105 Re: Help Me ! » Bottle with pills » 2016-01-11 09:55:54
Excellent work from all of you!! Many thanks for elaborating!!
#106 Re: Help Me ! » Bottle with pills » 2016-01-10 23:56:12
Guys thank you very much for your replies! Only by intuition, as I am not a programmer and I cannot write code (but fortunately I can understand a simple routine) this was also my reply. I experimented with 7 pills (4 black and 3 white) and I got 100% white. I guess that as the number increases, we get closer to 100%, provided that the number of black pills at the beginning is even.
Is there any way to prove this without the use of code (i.e. with the probabilities or by eliminating the possibility of a black pill at the end)?
#107 Help Me ! » Bottle with pills » 2016-01-09 23:36:18
- anna_gg
- Replies: 19
I was given an opaque bottle with 218 black and 217 white pills and I will play the following game with myself: I will draw 2 random pills and if they are both white, I keep one and put back the second in the bottle. If one is black and the other is white, I keep the white and return the black in the bottle. Finally, if both pills are black, I keep both of them and add a white pill in the bottle (from some extra stock that I keep separately). I keep playing until there is only one pill left in the bottle. What is the probability for the last pill to be white?
#108 Re: Help Me ! » Ice cream toppings » 2016-01-08 00:29:55
This is similar to the car plates problem, that must differ by one or two digits etc. I don't think I can solve that one either 
#109 Re: Help Me ! » 9 lockers » 2016-01-07 22:33:53
You must toggle the lockers by a total of 9+2x times (an odd number of moves), so that they are all turned open (9 moves) and 2x (that is, an even number) moves to open/close them. A workable solution would be 12345, 12367, 12389. This way we need 3 passes.
I agree for the second option (6 at a time): you cannot do it, because an odd number of moves (in total) is needed, but with 6 moves in each pass, we cannot get an odd number!
#110 Re: Help Me ! » 9 lockers » 2016-01-07 10:52:46
The final status must be OOOOOOOOO, not CCCCCCCCC
Theoretically five lockers should be doable with five passes. I don't know how yet but I'll try and find an example.
Six lockers can be done in three passes:
0: CCCCCCCCC 1: CCCOOOOOO 2: OOOOOOCCC 3: CCCCCCCCC
#111 Help Me ! » 9 lockers » 2016-01-06 22:33:31
- anna_gg
- Replies: 10
Suppose you are in a hallway lined with 9 closed lockers, which you want to open. In each pass, you must toggle exactly 5 lockers. What is the minimum number of passes will you need to open them all?
How about if you must toggle exactly 6 lockers in each pass?
#112 Re: Help Me ! » Ice cream toppings » 2016-01-06 21:22:35
I am trying to think of some other wording to help you understand Maybe the word "toppings" is confusing.
We have, for example:
1. Caramel sauce
2. M&Ms
3. Hot Fudge
4. Peanut butter
5. Oreo
6. Whipped cream
7. Strawberry sauce
...
and so on (14 different ingredients)
Each cone must have 4 of the above toppings (for example, Oreo, Strawberry sauce, Caramel Sauce and Peanut butter) but between ANY TWO ice-cream cones we cannot have more than ONE topping in common.
#113 Re: Help Me ! » Ice cream toppings » 2016-01-06 05:08:45
OK maybe I did not express it properly (sorry, english is not my native language):
Let's say we have infinite quantity of ice cream and cones and various toppings, also in adequate quantity each.
Let's name them T1, T2, T3... T14. One cone can have for example T1, T2, T5 and T12. Another can have T1, T3, T4 and T8 and another T2, T3, T9 and T11...
...and so on.
In other words, any two cones of ice cream cannot have more than one topping in common.
#114 Re: Help Me ! » Ice cream toppings » 2016-01-04 20:24:23
Just to clarify, there is an adequate quantity of each topping & ice cream and also infinite number of cones! (blood sugar levels skyrocket...)
#115 Re: Help Me ! » Ice cream toppings » 2016-01-04 09:09:17
Let alone with 4 toppings!!
I do not know where it is and besides ice cream is bad for your teeth.
#116 Re: Help Me ! » Ice cream toppings » 2016-01-04 09:08:11
No, order does not count.
Does the order of the toppings count? For instance, chocolate, sherbert, vanilla, strawberry as opposed to strawberry, chocolate, vanilla, sherbert.
#117 Help Me ! » Ice cream toppings » 2016-01-04 01:45:15
- anna_gg
- Replies: 46
An ice cream parlor has 14 different ice cream toppings. How many ice cream cones can we make, provided that we must place exactly 4 different toppings on each and also the cones must have at maximum one common topping with each other?
#118 Re: Help Me ! » Variation of mastermind game » 2014-08-14 01:02:23
Well, now that I reconsider the problem:
I agree that there are 27 permutations altogether. To be more specific, if we represent the 3 different colors with a, b and c, we would have:
aaa
aab
aac
aba
abb
abc
aca
acb
acc
baa
bab
bac
bba
bbb
bbc
bca
bcb
bcc
caa
cab
cac
cba
cbb
cbc
cca
ccb
ccc
As we can see, any combinations of 2 of the letters (out of 3), are repeated 3 times. For example, we have 3 x ?aa ?ab and ?ac, also 3 x b?a etc. Therefore with 9 tries at maximum, we can have a guaranteed win (2 colors in the correct place, thus 2 black pegs).
#119 Re: Help Me ! » Variation of mastermind game » 2014-08-13 19:51:40
Thanks Olinguito! So my understanding is that in the end, you do not subtract the winning sequence, right? It is just 27-6.
Can you please explain "Hence there are 27 − 6 − 1 = 20 permutations giving at most one black peg"?
#120 Help Me ! » Variation of mastermind game » 2014-08-12 19:00:59
- anna_gg
- Replies: 5
We are playing a mastermind game with 3 colors and 3 positions (holes). There is no white key peg (indicating the existence of a correct color code peg placed in the wrong position) - just black pegs, for each code peg from the guess that is correct in both color and position.
What is the minimum number of tries, to get at least 2 black pegs?
Duplicates are allowed.
#121 Re: Puzzles and Games » Electrical fuse board » 2014-08-12 06:53:07
Hmm now I see: If we had a string of length 2, the options would be YY, YN, NY (3 options).
For length 3, it would be YYY, YYN, YNY, NYY, NYN (5 options).
For length 4: YYYY, YYYN, YYNY, YNYY, YNYN, NYYY, NYYN and NYNY (8 options) and so on (13, 21 etc), thus for length 15 it would be 1597.
Thank you guys for your assistance!!
#122 Re: Puzzles and Games » Electrical fuse board » 2014-08-12 05:59:12
Hmmm... Fibonacci numbers...
#123 Re: Puzzles and Games » Electrical fuse board » 2014-08-12 03:54:14
How?
(I am neither a physicist nor a respectable mathematician, thus not contemptuous about proof...)
#125 Puzzles and Games » Electrical fuse board » 2014-08-12 02:23:07
- anna_gg
- Replies: 14
Hi everyone!
A computer data center is controlled by an electrical fuse board consisting of 15 fuses in one row.
The power in the data center goes off if any two consecutive fuses are in the OFF position (for example, if Y is ON and N is OFF, the power goes off if we have YYNNYYYYYYYYYYY but not if we have YNYNYNYNYNYYYYY). In how many different ways can we arrange the fuses, so as to always have power?