It is a continuous cycle and does not start again from the first person. So if there are 3 people, the count goes 1, 2(eliminated) 3, 1 (eliminated) and 3 is left.

Similarly for say 5 people it would be 1, 2(eliminated) 3, 4 (eliminated) 5, 1(eliminated), 3, 5(eliminated) and so #3 would be left.

Again, let us see if there is a pattern. We need a minimum of 2 people of course. Let’s try….
2 – 2
3 – 1
4 – 3
5 – 5
6 – 1
7 – 3
8 – 5
9 – 7
10 – 9
11 – 11
12 – 1
13 – 3
…
So do you see the pattern? Excepting the first case of 2 people, there is a clear pattern… Now I am sure you can figure out the rest from here…:)

Is the answer 217.. just to confirm ðŸ™‚ Thanks a lot for this technique

what’s the pattern here….if multiples of 3 wud give 1 then why not 9.

Sorry but I didn’t understand the statement “second last position of the person means second last winner”
As far as I understood,
If there are 3 person then 2nd person is second last winner(which is 2nd position)…So 1st will be the winner as you pointed out.
Same way for 4 person 3rd person will be second last winner(which is 3rd position)…So here again 1st person will be the winner…A bit confused here.

I’m sorry I could not make any sense of the comment. If we want to find the second last person, why are you finding the last one? You seem to be mixing up the two questions.

regards
J

Okay sir…I got it now…the game remains the same just instead of the last person we will check who was the last person to get eliminated…I checked various cases and it checks out…Thank You very much for the reply…I have so much gratitude for this blog and the people who came up with this noble idea.

What would be the general form of the pattern?
I am not able to figure that part out…but can used this to get to answer.
If N= Total number of knights
So,
1+[N-(Power of 2 X 3)]2 will be the 2nd last winner, where (Power of 2X3) has to be closest as well as less than N.
For ex in 300,
1+[300-(64X3)]2
1+[300-192]2
1+216
217th will be the runner up

Is the general form I have applied correct or it could be refined further?

As I said, you would have to figure out the rest, if at all you can develop pattern-recognition (if not this method is anyway useless)

regards
J

what will happen if numbers are removed in multiples of 3 or mulitiples of 4 or 5 ?

No simple solution in either of these two cases. You can try for small numbers and check if a pattern emerges, but the pattern turns out to be fairly complex in both the cases you mentioned….the latter case, maybe something in bases systems would be feasible, would have to think about it.

also what difference it will make when they are not removed continuously but at one time then started again ? for example in a count of 2000 number every 7 number is removed in every stage , what will be the last digit to be removed ? plz explain procedure

Very nasty, Gaurav. For 3 or 4 you could still hope to derive a kind of pattern by trying it for small numbers; for 10 we would have to do a lot of trials before a pattern would emerge (if any). Thus it would be an exercise for a computer program and not for a manual effort. Consequently, you could safely guarantee that it would be unlikely to come in a test/interview.

sir, in 1st question in 1st pass,
1,3,5,7,9… left
in second pass and in third pass also you started eliminating from 2nd no. in the row, then why did you chose 9 in 4th pass, shouldn’t it again be 1,17,25.. in 4th pass? and because the trick is also based on it. i got stuck here..

Because in the first 2 passes the last person eliminated was the last person left (#300 and #299 respectively). But in the third pass the last person eliminated was 293, so two steps further from there will be 297 (remains) and 1 (eliminated), i.e. 1 will have to be eliminated at the beginning of the 4th pass. I am not starting each pass from 1 again, it is a continuous pass as already explained in an earlier comment.

so can we conclude, that it is in the following a pattern of removing even terms(2nd, 4th, 6th, 8th..) in 1st three passes, odd ones in the next three passes and then evens and odds and so on in the gap of three passes?

sir,
when number of people are 4 first pass
1,2(eliminated),3,4(eliminated)
second pass
1(eliminated), #3 is left out?
could you please explain me how #1 was left as per your sol.

When there are 3 people, then how come the last remaining knight will be #3 ?

e.g. 1,2,3

1st pass: 1,3

2nd pass: 1

Please clarify.

Hi Nikunj,

It is a continuous cycle and does not start again from the first person. So if there are 3 people, the count goes 1, 2(eliminated) 3, 1 (eliminated) and 3 is left.

Similarly for say 5 people it would be 1, 2(eliminated) 3, 4 (eliminated) 5, 1(eliminated), 3, 5(eliminated) and so #3 would be left.

regards

J

Thank you. I appreciate it.

once i saw a question in which they asked about second last position of the person means second last winner . how to get that ???

Again, let us see if there is a pattern. We need a minimum of 2 people of course. Let’s try….

2 – 2

3 – 1

4 – 3

5 – 5

6 – 1

7 – 3

8 – 5

9 – 7

10 – 9

11 – 11

12 – 1

13 – 3

…

So do you see the pattern? Excepting the first case of 2 people, there is a clear pattern… Now I am sure you can figure out the rest from here…:)

regards

J

Is the answer 217.. just to confirm ðŸ™‚ Thanks a lot for this technique

what’s the pattern here….if multiples of 3 wud give 1 then why not 9.

Sorry but I didn’t understand the statement “second last position of the person means second last winner”

As far as I understood,

If there are 3 person then 2nd person is second last winner(which is 2nd position)…So 1st will be the winner as you pointed out.

Same way for 4 person 3rd person will be second last winner(which is 3rd position)…So here again 1st person will be the winner…A bit confused here.

I’m sorry I could not make any sense of the comment. If we want to find the second last person, why are you finding the last one? You seem to be mixing up the two questions.

regards

J

Okay sir…I got it now…the game remains the same just instead of the last person we will check who was the last person to get eliminated…I checked various cases and it checks out…Thank You very much for the reply…I have so much gratitude for this blog and the people who came up with this noble idea.

What would be the general form of the pattern?

I am not able to figure that part out…but can used this to get to answer.

If N= Total number of knights

So,

1+[N-(Power of 2 X 3)]2 will be the 2nd last winner, where (Power of 2X3) has to be closest as well as less than N.

For ex in 300,

1+[300-(64X3)]2

1+[300-192]2

1+216

217th will be the runner up

Is the general form I have applied correct or it could be refined further?

As I said, you would have to figure out the rest, if at all you can develop pattern-recognition (if not this method is anyway useless)

regards

J

what will happen if numbers are removed in multiples of 3 or mulitiples of 4 or 5 ?

No simple solution in either of these two cases. You can try for small numbers and check if a pattern emerges, but the pattern turns out to be fairly complex in both the cases you mentioned….the latter case, maybe something in bases systems would be feasible, would have to think about it.

regards

J

also what difference it will make when they are not removed continuously but at one time then started again ? for example in a count of 2000 number every 7 number is removed in every stage , what will be the last digit to be removed ? plz explain procedure

Dear J, i’m able to solve a lot of other variants of this question. Thank you so much. ðŸ™‚

thanks sir, i am now able to figure out where i was going wrong when i solved a logical question of the same pattern. ðŸ˜€

HI Sir,

How about if lets say every 10th was to be eliminated?

Very nasty, Gaurav. For 3 or 4 you could still hope to derive a kind of pattern by trying it for small numbers; for 10 we would have to do a lot of trials before a pattern would emerge (if any). Thus it would be an exercise for a computer program and not for a manual effort. Consequently, you could safely guarantee that it would be unlikely to come in a test/interview.

regards

J

sir, in 1st question in 1st pass,

1,3,5,7,9… left

in second pass and in third pass also you started eliminating from 2nd no. in the row, then why did you chose 9 in 4th pass, shouldn’t it again be 1,17,25.. in 4th pass? and because the trick is also based on it. i got stuck here..

Because in the first 2 passes the last person eliminated was the last person left (#300 and #299 respectively). But in the third pass the last person eliminated was 293, so two steps further from there will be 297 (remains) and 1 (eliminated), i.e. 1 will have to be eliminated at the beginning of the 4th pass. I am not starting each pass from 1 again, it is a continuous pass as already explained in an earlier comment.

regards

J

so can we conclude, that it is in the following a pattern of removing even terms(2nd, 4th, 6th, 8th..) in 1st three passes, odd ones in the next three passes and then evens and odds and so on in the gap of three passes?

NO. Every different starting number of people will have a different sequence of passes.

regards

J

sir,

when number of people are 4 first pass

1,2(eliminated),3,4(eliminated)

second pass

1(eliminated), #3 is left out?

could you please explain me how #1 was left as per your sol.

You have eliminated 4 and then 1…that is, two consecutive people! After 4 is kicked out, 1 must stay and hence 3 will get kicked out.

regards

J

I couldn’t understand the question first