This is a very interesting problem, constraints are within problem descriptions XD. I thought there should be no solution that satisfies all constraints. But there indeed is..
Original Problem
I met this problem on a message on Wechat (only 4 options each), then I found a harder (maybe original) problem here:
1. The first question that have answer (b) is:
(a) 2, (b) 3, (c) 4, (d) 5, (e) 6
2. The only continous questions that have the same answer are:
(a) 2 & 3, (b) 3 & 4, (c) 4 & 5, (d) 5 & 6, (e) 6 & 7
3. The answer of this question is equal to the answer of which question?:
(a) 1, (b) 2, (c) 4, (d) 7, (e) 6
4. How many questions have the answer as (a)?
(a) 0, (b) 1, (c) 2, (d) 3, (e) 4
5. The answer of this question is equal to the answer of which question?:
(a) 10, (b) 9, (c) 8, (d) 7, (e) 6
6. The number of answer of (a) is equal to the number of which answer?:
(a) b, (b) c, (c) d, (d) e, (e) none of the above
7. The difference between the answer of this question and next question is:
(The difference between a & b is 1)
(a) 4, (b) 3, (c) 2, (d) 1, (e) 0
8. How many answers are vowels (a,e)?
(a) 2, (b) 3, (c) 4, (d) 5, (e) 6
9. The number of answers that is non-vowels (b,c,d) is?:
(a) prime number, (b) X!, (c) x^2, (d)x^3 , (e) 5x
10.The answer of this question is:
(a) a, (b) b, (c) c, (d) d, (e) e
Interesting, huh!!
SPOILER ALERT
My solution and the official solution are listed below, do not “accidentally” see it if you want to give it a try!
My Solution
I assume you have tried by yourself or you are just eager to know my thoughts, so here is what I have done:
Cut impossible options
(b) in 1, (a) in 2, (a) in 4, (d) in 8, (e) in 9 are impossible options, see why?
- (b) in 1: if 3 is (b), then 2 is (b), but (a) in 2 is also correct!
- (a) in 2: if 2 is (a), then 3 is (a), then 1 is (a), then 2 is (b)!
- (a) in 4: if 4 is (a), then there exists some (a)
- (d) in 8 && (e) in (9): list out 1 to 10, which can be prime or factorial or square or cube or 5divided?
If some number satisfies more than one option, it is definitely impossible. Otherwise, which option should we choose?
In this way, prime => 7, factorial => 6, square => 4, cube => 8.
My first thought was to use 3 and 5 as the key to solution. In my opinion, there should NEVER be any ambiguity.
In other words, if one option is selected as the answer for a question, all other options must be FALSE!
In this way, through 3 and 5 we can eliminate many impossible permutations.- 3:A, 5:any – failed
- 3:B, 5:any – failed
- 3:C, 5:A – failed
- 3:C, 5:B – failed
- 3:C, 5:C – failed (D/E are impossible)
- 3:D, 5:D – failed (A/B/C/E are impossible)
- 3:E, 5:E (A/B/C/D are impossible, so this must be the answer! We are almost done!)
- 3:E, 5:E, 8:E – failed
- 3:E, 5:E, 8:B – failed (only B/E is possible for 8)
Whaaaaat? all failed?! Below is what I got this when assigning 3:E, 5:E
1. The first question that have answer (b) is: (c) 4 2. The only continous questions that have the same answer are: (d) 5 & 6 3. The answer of this question is equal to the answer of which question?: (e) 6 4. How many questions have the answer as (a)? (b) 1 5. The answer of this question is equal to the answer of which question?: (e) 6 6. The number of answer of (a) is equal to the number of which answer?: (e) none of the above 7. The difference between the answer of this question and next question is: (The difference between a & b is 1) (a) 4, (b) 3, (c) 2 8. How many answers are vowels (a,e)? (b) 3, (e) 6 9. The number of answers that is non-vowels (b,c,d) is?: (a) prime number => 7, (c) x^2 => 4 10. The answer of this question is: (b) b, (c) c, (d) d, (e) e
The Solution
Here is that “solution”:
1. (c)
2. (d)
3. (e)
4. (b)
5. (e)
6. (e) // till 6, we have the same answer
7. (d) // but for 7, in my opinion, how can it be (d)?!
// If so, 5 can also choose (d), AMBIGUITY arises!!
8. (c)
9. (b)
10.(a)
In fact, my hypothesis that there should be no ambiguity may not be stable.
But I think everyone would agree that the problem will be better if there is indeed no ambiguity at all.
I am still glad that I have solved most (is ≥60% “most”?) of it!!