Self Referential Weekender.
I like this type of quizzes. This one I have copied from a site and the original author is Martin Henz. The solution is available on net but it is great fun to solve. The original url is polaris.lcc.uma.es/~afdez/srq/srq or something but is very slow. So I will reproduce the questions here.
A good exercise would be to design another quiz like this yourself that has a unique solution set. Have fun.
1. The first question whose answer is A is question
(A) 4 (B) 3 (C) 2 (D) 1 (E) none of the above
2. The only two consecutive questions with identical answers are questions
(A) 3 and 4 (B) 4 and 5 (C) 5 and 6 (D) 6 and 7 (E) 7 and 8
3. The next question with answer A is question
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
4. The first even numbered question with answer B is question
(A) 2 (B) 4 (C) 6 (D) 8 (E) 10
5. The only odd numbered question with answer C is question
(A) 1 (B) 3 (C) 5 (D) 7 (E) 9
6. A question with answer D
(A) comes before this one, and not after this one
(B) comes after this one, and not before this one
(C) comes before and after this one
(D) does not occur at all
(E) none of the above
7. The last question whose answer is E is question
(A) 5 (B) 6 (C) 7 (D) 8 (E) 9
8. The number of questions whose answers are consonants is
(A) 7 (B) 6 (C) 5 (D) 4 (E) 3
9. The number of questions whose answers are vowels is
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4
10.The answer to this question is
(A) A (B) B (C) C (D) D (E) E
A good exercise would be to design another quiz like this yourself that has a unique solution set. Have fun.
1. The first question whose answer is A is question
(A) 4 (B) 3 (C) 2 (D) 1 (E) none of the above
2. The only two consecutive questions with identical answers are questions
(A) 3 and 4 (B) 4 and 5 (C) 5 and 6 (D) 6 and 7 (E) 7 and 8
3. The next question with answer A is question
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
4. The first even numbered question with answer B is question
(A) 2 (B) 4 (C) 6 (D) 8 (E) 10
5. The only odd numbered question with answer C is question
(A) 1 (B) 3 (C) 5 (D) 7 (E) 9
6. A question with answer D
(A) comes before this one, and not after this one
(B) comes after this one, and not before this one
(C) comes before and after this one
(D) does not occur at all
(E) none of the above
7. The last question whose answer is E is question
(A) 5 (B) 6 (C) 7 (D) 8 (E) 9
8. The number of questions whose answers are consonants is
(A) 7 (B) 6 (C) 5 (D) 4 (E) 3
9. The number of questions whose answers are vowels is
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4
10.The answer to this question is
(A) A (B) B (C) C (D) D (E) E
posted by Nikhil at 8:28 AM
8 Comments:
1. C
2. A
3. B
4. B
5. A
6. B
7. E
8. B
9. E
10. D
That is what I got and I doubt that there are two ways to do it (The original page said that it does not.)
I started doing this ... and really it was to just Assume 1) A ... run through the rest of the cascading answers to see whether the answers loop back.
Once I figured it was just a matter of computational cycles, I lost interest :)
Is there another way?
Yes, there is another way. The way I did it, I could find that there are only two choices possible for #1.
I do not have my sheet with me, so let me tell what I still remember.
For #8 and #9, the addition has to be 10, so the only possible answer pairs are 'A and D' or 'B and E'.
For #1, it cannot be D as it would be self-contradicting.
Also, you can prove that two more for #1 are not possible. (you need not run through ALL other questions. You get a contradiction in 2-3 steps.That way you can cancel out some other options too. )
Once you narrow down the options for #1, it is relatively simpler.
I will post the complete sequence of steps that I used once this dies down. I really need my sheet for that :)
And Axe, A for #1 is self-contradicting also. You can get rid of one more option in 2-3 steps.
Hi Nikhil!
Nice blog! Seems to have quite a bit of traffic on it.
Btw, tere post ke title mein ek typo hai :-)
Cheers,
Anand
What typo Anand? :)
Mail me your contact info. Are you planning to attend Abhijeet's reception in Nagpur?
I just saw your comments on the "Friends ?" problem in Sushrut's blog. http://www.blogger.com/comment.g?blogID=9869106&postID=112373715711355983
Your last sentence "Either..." is not correct. You would definitely have lost some marks there ;-)
Also, the following proof is neater than yours or Abhijeet's. The problem is essentially to show that in any undirected graph, there are two vertices of the same degree.
Assuming number of vertices >= 2, either there are two vertices of zero degree or there is a connected component C of size at least 2. Now, we will show that the property holds for C. Since C is connected, degree of each vertex in C >= 1. Further, degree can be at most n-1, where n is the size of the connected component. Since there are n vertices and n-1 possible distinct values their degrees can take, two of them must have same degree.
Q: Obviously, the same property holds for an undirected tree. Can you state something stronger for an undirected tree though?
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