# puzzle question

Watch
Announcements
#1
I know by doing 2,2,0 =2 and 6,6,6=18. This gives the lowest and highest possible scores.
Therefore, there are 17 possible numbers to exclude. But this is a really time-consuming method. Is there an easier way to identify which numbers can and can't be made?
2
3
4
5
6
7
8
9
10
11
12
13
14
15
17
18
Last edited by As.1997; 1 year ago
0
1 year ago
#2
(Original post by As.1997)
I know by doing 2,2,0 =2 and 6,6,6=18.
Therefore, there are 17 possible numbers to exclude. But this is a really time-consuming method. Is there an easier way to identify which numbers
18
If you consider all 3 darts landing in circles, then since the numbers are 2,4,6, it's pretty clear, to my mind, that you can get all even numbers from 6 to 18. I.e. 7 in all.

Then if you get 2 in circles and one miss, then it effectively divides the numbers by 2, so you're hitting 2 from 1,2,3, which give you all values from 2 to 6 inclusive. I.e. 5 numbers.

So, total number is 7+5 and minus 1 as 6 is in both sets, giving 11, or A. Can't see how it could be C, unless I misinterpreted the question.
Last edited by ghostwalker; 1 year ago
0
#3
(Original post by ghostwalker)
If you consider all 3 darts landing in circles, then since the numbers are 2,4,6, it's pretty clear, to my mind, that you can get all even numbers from 6 to 18. I.e. 7 in all.

Then if you get 2 in circles and one miss, then it effectively divides the numbers by 2, so you're hitting 2 from 1,2,3, which give you all values from 2 to 6 inclusive. I.e. 5 numbers.

So, total number is 7+5 and minus 1 as 6 is in both sets, giving 11, or A. Can't see how it could be C, unless I misinterpreted the question.
I found this explanation. The approach they took is to literally go through every single number, which is what I did but eventually, I realised this will take me forever so I just skipped the question.
I guess one way to save time is to realise since 2,4,6 are even numbers, chances are all even numbers upto 18 can be made. So the next thing would be to eliminate the odd numbers.
Leaving us: 3, 5, 7, 9, 11, 13, 15, 17
(it claims 11, 13, 15, 17 are impossible to make -- I find this a little weird since I would only know this if I tried several combinations).
Last edited by As.1997; 1 year ago
0
1 year ago
#4
(Original post by As.1997)
I found this explanation. The approach they took is to literally go through every single number, which is what I did but eventually, I realised this will take me forever so I just skipped the question.
I see where I went wrong. For two hits I was halving the total score, rather than the score so far.

In which case:

As before 3 hits, gives you 6 through 18, evens only.

Two hits and then a miss, gives you 2 through 6, as before, bringing the total to 11.

And finally hit, miss hit, gives:
First hit then miss gives 1,2, or 3. And another hit of 2,4, or 6, gives 3,4,5 or 5,6,7, or 7,8,9. And the only numbers there we don't have already are 7 and 9. So, 2 more bringing the total to 13 (C).

Note, Miss, hit, hit, would give you evens from 4 to 12, which we have already.
0
#5
(Original post by ghostwalker)
As before 3 hits, gives you 6 through 18, evens only.
2,2,2= 6
6,6,6=18
(I assumed with 3 hits since the lowest and highest are 6 and 18 and the numbers that can be hit are (2,4,6 --i.e. even) then all even numbers between 6 through 18 must be attainable, is this what you also assumed?)

With two hits and a miss, the lowest and highest I can get are:
2,2,0=2
0,6,6=12
So here, I would go wrong to assume I can get all numbers from 2 through 12.
Last edited by As.1997; 1 year ago
0
1 year ago
#6
(Original post by As.1997)
2,2,2= 6
6,6,6=18
(I assumed with 3 hits since the lowest and highest are 6 and 18 and the numbers that can be hit are (2,4,6 --i.e. even) then all even numbers between 6 through 18 must be attainable, is this what you also assumed?)
Didn't need to assume it - it seems obvious as 2,4,6 goes up in 2's.

(Original post by As.1997)
With two hits and a miss, the lowest and highest I can get are:
2,2,0=2
0,6,6=12
So here, I would go wrong to assume I can get all numbers from 2 through 12.
True, you would go wrong.
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (1)
100%
No (0)
0%