# puzzle question

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Please refer to the attachment.

The answer is C.

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

The answer is C.

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

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#2

(Original post by

Please refer to the attachment.

The answer is C.

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

**As.1997**)Please refer to the attachment.

The answer is C.

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

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

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(Original post by

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.

**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 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

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#4

(Original post by

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.

**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.

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.

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(Original post by

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

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

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

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#6

(Original post by

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?)

**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?)

(Original post by

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.

**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.

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