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#76 Re: Help Me ! » Skippy the kangaroo » 2016-02-14 00:37:11
No, because, in all 3 cases, the 7 meters are reached before he takes the backward jump.
Hi Anna;
Please let me know if these three jump sequences - expressed in {1,2,-1} terms - enable Skippy to reach the bowl:
A. 1,1,1,1,1,1,1,-1,1
B. 2,1,1,2,1,-1,1
C. 2,2,1,2,-1,1
I think this is it, but without adding f(7) in the end; thus 146.
The calculations in detail.
[list=*]
[/*]
[*]
[/list]Here f(n) is the number of ways Skippy can skip a distance of exactly n metres without moving backwards. If his initial skip is 1 metre, he has n−1 metres left to skip; if his starts by skipping 2 metres, he has n−2 metres left to do.
For backwardness:
[list=*]
[/*]
[*]
[/list]g(m) is the number of ways to get to the bowl from 7 metres away, skipping back a metre at the m-metre mark from the starting point. The kangaroo can skip the first m metres in f(m) ways; after the backward skip, he will be 8−m metres from his bowl, and can get there in f(8−m) ways.
Hence the total number of ways to skip to his bowl from 7 metres away (assuming he has the option to decline skipping backwards) is
[list=*]
[/*]
[*]
[/list]
#77 Re: Help Me ! » Skippy the kangaroo » 2016-02-13 20:44:50
A great job by all of you. Indeed the backwards jump is mandatory and Skippy cannot reach the bowl without having taken the back jump (i.e. after he reaches the 7 meters). I therefore start to believe that 146 is correct.
#78 Re: Help Me ! » Skippy the kangaroo » 2016-02-12 12:01:59
Why did you change your mind? I also got 167.
Sorry, change of mind.
#79 Re: Help Me ! » 5 students test » 2016-02-11 20:23:00
Unfortunately not 
But I will definitely work more on this riddle and will post any update. Thanks for your very valuable input!
Hi Anna,
If you know the mathematical formula for the solution, would you please post it? Thanks.
#80 Re: Help Me ! » 5 students test » 2016-02-11 08:49:52
Correct!! Very well done!
#81 Help Me ! » Rotten apples » 2016-02-11 01:19:32
- anna_gg
- Replies: 22
We have a basket with 8 apples of which 2 are rotten; one is lighter than the good apples and the second, strangely enough, is heavier; maybe because of the worms that have penetrated it 
All the good apples are of the same weight.
How can we tell if the two rotten apples together weight more or equal or less than two good ones - in only 3 weightings with a two-arm balance scale? Obviously there are no visible signs of the rotten apples.
#82 Re: Help Me ! » Skippy the kangaroo » 2016-02-10 18:34:36
Right, and this cannot be the first jump (i.e. Skippy cannot start by jumping backwards). I also did a correction in the wording, to explain that it cannot jump past to the bowl and then return with a backward jump; it must arrive exactly to the bowl with a forward jump (the last one).
#83 Help Me ! » Skippy the kangaroo » 2016-02-10 10:46:07
- anna_gg
- Replies: 42
Skippy the kangaroo is standing 7 meters away from his feeder bowl and wants to reach it. It can jump one or more jumps ahead and each jump can be of one or two meters long, and also one jump backwards, which is one meter long. Skippy can reach its bowl by a sequence of one or more jumps ahead and only one jump backwards, but not the first one. In how many different ways can Skippy reach its bowl (consider the bowl as a point without dimensions and also that it must arrive exactly at the edge of the bowl without bypassing it - as if it is placed in front of a wall)?
#84 Re: Help Me ! » Square ceiling » 2016-02-10 10:16:12
Genius!!
From the way I've spaced the circles vertically, the image shows the top row of lights right at the top of the square. However, if spaced at the minimum amount there is a vacant strip of 4√0.75 across the top. Too small to do anything with, I suppose...but I did have a look to see if I could combine it somehow with other unused spaces to conjure up another spotlight, just in case this puzzle came from some sneaky puzzle setter!
For that I tried diamond shapes and equilateral triangles, but gave up after a while.
#85 Re: Help Me ! » Password lock » 2016-02-04 10:06:28
Phrontister, you are a genius!!!
Many thanks to all of you, though!
#86 Re: Help Me ! » Password lock » 2016-02-03 03:30:34
I think Phrontister's strategy in post #6 is a good starting point, but I believe (though I cannot prove it!!) that it can be done in 5 guesses, if you pick some other sets instead of the triplets. Maybe this way we manage to eliminate all 6 of the remaining combinations.
Can anyone write a program to test all possible combinations of 5 guesses (from the total of 27) - I think they are 80730 - to see which 5 will finally open the lock? Of course, I may be wrong
#87 Re: Help Me ! » Password lock » 2016-01-31 08:04:59
Right, all 7 combinations will open the lock.
In other words, if the password is 123, then all of the following will open the lock, right?
[list=*]
[*]123, 223, 323, 113, 133, 121, 122[/*]
[/list]
My question was whether the wildcard could be in any position, or whether it was in a definite (but unknown) position. In the former case, the lock will open for any of the above seven combinations; in the latter case, then, for example, if the wildcard is in the centre, only 113, 123 and 133 will open the lock.
#88 Re: Help Me ! » Password lock » 2016-01-31 04:01:16
Any of them, but the "any" (asterisk or wildcard) digit must be in the correct position. For example 231 won't open the lock, but 223 will.
anna_gg wrote:For example, if the password is 123, the lock opens when you try 1*3, or *23, or 12*.
Do you mean any one of 1*3, *23, 12* can open the lock, or just one of them will?
If any one of them works,
#89 Help Me ! » Password lock » 2016-01-30 21:48:54
- anna_gg
- Replies: 23
There is a lock with the combination between 111-333 (3 digits, each with values 1, 2 or 3, duplicates allowed, of course).
The lock is defective and we could use only two digits to open the lock. For example, if the password is 123, the lock opens when you try 1*3, or *23, or 12*. Compose a strategy to find the minimum number of tries to open the lock, under the worst situation and prove that this strategy is optimal.
#90 Re: Help Me ! » Square ceiling » 2016-01-30 08:29:12
It seems we can put more...
Why is that?
#91 Re: Help Me ! » Square ceiling » 2016-01-29 09:04:04
It doesn't seem to be correct
Hi,
I found a way to fit 17 lights. There is enough space available for 18, but I would be astonished if it was usable.
I'm not really sure how to describe the (my) solution without an image. Basically:
There is a light at each corner.
Two of the corner lights opposite each other have 1.5m gaps on either side, and then there are three lights in a row 1m apart on each side (including the other corner).
There is a light as close as possible to both corner lights with a gap on either side. It looks like these lights are sqrt(2)/2 metres up/down and sqrt(2)/2 metres across from the corner lights, but that is just a guess based on physical measurement.
There is a light as far across as possible from the two corner section lights, i.e. 1m across and 1m up from one of the other corners.
Finally, there is room for two more lights as close as possible to each corner section light (those lights have a calculable distance that is about 2.95m)
#92 Re: Help Me ! » 5 students test » 2016-01-29 01:20:11
Yes, this is such an example!!
Hi anna_gg,
anna_gg wrote:In how many possible ways can this be done?
Is this an example of two different ways?
1 2 A: 1,3 1,3 B: 1,4 1,3 C: 2,4 2,4 D: 2,5 2,5 E: 3,5 4,5It seems so to me, and my answer in post #10 is based on this example being valid.
#93 Re: Help Me ! » 5 students test » 2016-01-24 23:13:25
Guys, thanks for your replies; however, I am not really sure I have correctly explained the requirement: Each student (A, B, C, D and E) must pick exactly 2 of the 5 questions, for example A can choose only 1 and 2, B can choose 2 and 3 but then, question Nr 2 cannot be chosen by any other student (as it has already been chosen by A and B) and so on. This means that there is no restriction for the repetition of digits in each set: we can have for example ABCDE -->>11223 (both A and B picked question 1).
#94 Help Me ! » Square ceiling » 2016-01-23 01:26:41
- anna_gg
- Replies: 11
On a square ceiling 3.5 x 3.5 meters we want to place spot lights with minimum 1 m distance from each other. What is the maximum number of lights we can place?
#95 Re: Help Me ! » License plates in Riddleland » 2016-01-22 21:18:22
@Phrontister: It seems your result is correct, but through some other solution
For the first 7 digits, obviously the maximum number of possible different plates with at least one different digit in one position (given that each digit can have any value from 0 to 9 and repetition IS permitted), is 10^7.
Now we must find a way to add an 8th digit in such a way that each plate is different from each other also in the last digit. That is, if we consider two (7-digit) plates that only differ in 1 of the 7 digits, the 8th must be selected in such a way that it always creates a different 8-digit plate. We are therefore looking for a checksum - for example calculating the sum of the first 7 digits --> MOD(10) and then selecting the last digit to be -MOD(10), so that the sum of all 8 digits is always a product of 10.
So in effect, yes, the maximum number is 10^7!
#96 Re: Help Me ! » License plates in Riddleland » 2016-01-18 11:35:34
Unfortunately not!
#97 Re: Help Me ! » 5 students test » 2016-01-17 08:48:24
Could be; I don't know the answer. Solution please?
#98 Help Me ! » 5 students test » 2016-01-16 04:14:43
- anna_gg
- Replies: 28
5 students are to pass a graduation exam and for this purpose they must pick 2 questions from a pool of 5.
In fact, each student must choose exactly 2 questions and each question must be chosen by exactly 2 students. In how many possible ways can this be done?
#99 Re: Help Me ! » License plates in Riddleland » 2016-01-15 10:16:48
Repetition of digits is allowed.
#100 Help Me ! » License plates in Riddleland » 2016-01-14 10:27:01
- anna_gg
- Replies: 7
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?