12
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There are lots of easy questions:

How many colors are in a rainbow? How much do 2018 pennies cost? Why did I add an extra letter to 'false'?

However, this riddle, is not one of those.

You are to say which of these are true, and which are false.

If a statement is not 100% right, it's false.

Good luck.

  1. Number 7 and 8 are false.

2. Four is false, twelve is true.

  1. Number 20 is true.

  2. One is true, two is false.

5. 3, 9, and 6, all are false.

  1. I don't like strawberries.

  2. Number 13 is true.

8. I like strawberries.

  1. Number 5 is false.

  2. Number 14 is true.

  3. Number 11 is true, and 6 is, too.

  4. 20 is false.

13. Number 1 and 8 are false.

  1. Number 17 and 12 and 13 are false.

  2. This statement is true, and so is statement 2.

  3. The preceding statements are more true than false.

  4. Either 11 or 15 is false, but 19 is definitely true.

  5. Five and twelve are true.

  6. There are more than one correct answer(s).

  7. Half of the questions are false.

  8. 9 is false. 2 is true. 1 is true, if 6 is, too.

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  • $\begingroup$ Does the bolded second statement mean anything? $\endgroup$
    – 1848
    Commented Aug 18, 2018 at 0:42
  • 2
    $\begingroup$ They're distractions. Just find which statements are true, and which are false. $\endgroup$
    – Alto
    Commented Aug 18, 2018 at 1:02
  • 2
    $\begingroup$ That's what #19 is asking. $\endgroup$
    – Alto
    Commented Aug 18, 2018 at 1:14
  • 1
    $\begingroup$ What's "half" of 21? Is it 10 or 11? (rot13: Be qbrf gung zrna 20 vf nhgbzngvpnyyl snyfr?) $\endgroup$
    – sedrick
    Commented Aug 18, 2018 at 8:21
  • 1
    $\begingroup$ why the bolding, spoilering, italicizing, and strikethrough? $\endgroup$
    – Kate Gregory
    Commented Aug 18, 2018 at 13:51

3 Answers 3

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Long solution (not sure if there's a more clever way without using a computer):

I will start with an assumption:

Since 21 is an odd number, question 20 is automatically false.

I will strikethrough false statements and bold true statements.

What we currently know:

1. Number 7 and 8 are false.
2. Four is false, twelve is true.
3. Number 20 is true.
4. One is true, two is false.
5. 3, 9, and 6, all are false.
6. I don't like strawberries.
7. Number 13 is true.
8. I like strawberries.
9. Number 5 is false.
10. Number 14 is true.
11. Number 11 is true, and 6 is, too.
12. 20 is false.
13. Number 1 and 8 are false.
14. Number 17 and 12 and 13 are false.
15. This statement is true, and so is statement 2.
16. The preceding statements are more true than false.
17. Either 11 or 15 is false, but 19 is definitely true.
18. Five and twelve are true.
19. There are more than one correct answer(s).
20. Half of the questions are false.
21. 9 is false. 2 is true. 1 is true, if 6 is, too.

Then, we answer some more questions based on the new questions we were able to answer:

1. Number 7 and 8 are false.
2. Four is false, twelve is true.
3. Number 20 is true.
4. One is true, two is false.
5. 3, 9, and 6, all are false.
6. I don't like strawberries.
7. Number 13 is true.
8. I like strawberries.
9. Number 5 is false.
10. Number 14 is true.
11. Number 11 is true, and 6 is, too.
12. 20 is false.
13. Number 1 and 8 are false.
14. Number 17 and 12 and 13 are false.
15. This statement is true, and so is statement 2.
16. The preceding statements are more true than false.
17. Either 11 or 15 is false, but 19 is definitely true.
18. Five and twelve are true.
19. There are more than one correct answer(s).
20. Half of the questions are false.
21. 9 is false. 2 is true. 1 is true, if 6 is, too.

At this point, we do casework

Case 1: when Number 1 is true
Case 2: when Number 1 is false.

Case 1: (This is quite easy to fill up so I'll skip some steps.)

1. Number 7 and 8 are false.
2. Four is false, twelve is true.
3. Number 20 is true.
4. One is true, two is false.
5. 3, 9, and 6, all are false.
6. I don't like strawberries.
7. Number 13 is true.
8. I like strawberries.
9. Number 5 is false.
10. Number 14 is true.
11. Number 11 is true, and 6 is, too.
12. 20 is false.
13. Number 1 and 8 are false.
14. Number 17 and 12 and 13 are false.
15. This statement is true, and so is statement 2.
16. The preceding statements are more true than false.
17. Either 11 or 15 is false, but 19 is definitely true.
18. Five and twelve are true.
19. There are more than one correct answer(s).
20. Half of the questions are false.
21. 9 is false. 2 is true. 1 is true, if 6 is, too.

Now

11. Number 11 is true, and 6 is, too.
The statement "This statement is true" can be both true and false. Since 6 is true, then 11 can be both true or false. This means there's more than 1 possible answer.

Furthermore,

2. Four is false, twelve is true.
4. One is true, two is false.
15. This statement is true, and so is statement 2.

Since 12 is true and 1 is true, then statements 2 and 4 are basically
2. 4 is false.
4. 2 is false

In this case, either one is true and the other will be false. Also, number 15 = number 2.

From the two points above, we have the following possibilities:

11 is true. 2 is true and 4 is false. 15 is true.
11 is true. 2 is false and 4 is true. 15 is false.
11 is false. 2 is true and 4 is false. 15 is true.
11 is false. 2 is false and 4 is true. 15 is false.

From these, 16, 17, and 19 should come easily.

We get the following possible true statements for case 1:

[1, 6, 9, 12, 19, 2, 15, 11]
[1, 6, 9, 12, 19, 2, 15, 17]
[1, 6, 9, 12, 19, 4, 11, 17]
[1, 6, 9, 12, 19, 4, 17]

Proceeding to case 2 (once again skipping some steps):

1. Number 7 and 8 are false.
2. Four is false, twelve is true.
3. Number 20 is true.
4. One is true, two is false.
5. 3, 9, and 6, all are false.
6. I don't like strawberries.
7. Number 13 is true.
8. I like strawberries.
9. Number 5 is false.
10. Number 14 is true.
11. Number 11 is true, and 6 is, too.
12. 20 is false.
13. Number 1 and 8 are false.
14. Number 17 and 12 and 13 are false.
15. This statement is true, and so is statement 2.
16. The preceding statements are more true than false.
17. Either 11 or 15 is false, but 19 is definitely true.
18. Five and twelve are true.
19. There are more than one correct answer(s).
20. Half of the questions are false.
21. 9 is false. 2 is true. 1 is true, if 6 is, too.

Similar to case 1:

15 is like the 11 of case 1. It can be both true or false.

Moreover:

5 and 9 are like the 2 and 4 of case 1: one is true and the other is false.

If 5 is true, 9 is false. 6 is false. 8 is true. 11 is false, 17 is true, 18 is true, 13 is false, 7 is false, 21 is true.

If 5 is false, 9 is true, 21 is false, 18 is false.
Subcase 1: If 7 is true, then 13 is true and 8 is false and 6 is true and 11 can be true/false.
Subcase 2: If 7 is false, then 13 is false and 8 is true and 6 is false and 11 is false and 17 is true.

Concluding, case 2 gives us:

[2, 12, 19, 5, 8, 17, 18, 21, 15]
[2, 12, 19, 5, 8, 17, 18, 21]
[2, 12, 19, 9, 7, 13, 6, 11, 15]
[2, 12, 19, 9, 7, 13, 6, 11, 17]
[2, 12, 19, 9, 7, 13, 6, 15, 17]
[2, 12, 19, 9, 7, 13, 6]
[2, 12, 19, 9, 8, 17]

So I believe the total is

[1, 2, 6, 9, 11, 12, 15, 19]
[1, 2, 6, 9, 12, 15, 17, 19]
[1, 4, 6, 9, 11, 12, 17, 19]
[1, 4, 6, 9, 12, 17, 19]
[2, 5, 8, 12, 15, 17, 18, 19, 21]
[2, 5, 8, 12, 17, 18, 19, 21]
[2, 6, 7, 9, 11, 12, 13, 15, 19]
[2, 6, 7, 9, 11, 12, 13, 17, 19]
[2, 6, 7, 9, 12, 13, 15, 17, 19]
[2, 6, 7, 9, 12, 13, 19]
[2, 8, 9, 12, 17, 19]

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  • 1
    $\begingroup$ F*** brute-force — this is what puzzling is all about: using the mind. I just spent my last upvote on an answer of a cipher puzzle unfortunately... so now I have to wait $12$ hours before I can upvote again. Nonetheless, this is a brilliant answer, whether it is wrong or right :D $\endgroup$
    – Mr Pie
    Commented Aug 18, 2018 at 11:21
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    $\begingroup$ My understanding of Q17 ("Either 11 or 15 is false") is that both 11 and 15 cannot be false for 17 to be true (XOR). This eliminates [1, 6, 9, 12, 19, 4, 17], [2, 12, 19, 5, 8, 17, 18, 21] and [2, 12, 19, 8, 17]. $\endgroup$
    – xhienne
    Commented Aug 18, 2018 at 19:52
  • $\begingroup$ I interpreted as OR instead of XOR, but yeah reading it again, the XOR interpretation indeed makes more sense. $\endgroup$
    – sedrick
    Commented Aug 18, 2018 at 21:20
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A brute-force solution:

import itertools

def satisfied(Q):
    return all([
        Q[1] == ((not Q[7]) and (not Q[8])),
        Q[2] == (not Q[4]) and Q[12],
        Q[3] == (Q[20]),
        Q[4] == (Q[1] and (not Q[2])),
        Q[5] == ((not Q[3]) and (not Q[9]) and (not Q[6])),
        Q[7] == (Q[13]),
        Q[8] == (not Q[6]),
        Q[9] == (not Q[5]),
        Q[10] == Q[14],
        Q[11] == (Q[11] and Q[6]),
        Q[12] == (not Q[20]),
        Q[13] == ((not Q[1]) and (not Q[8])),
        Q[14] == ((not Q[17]) and (not Q[12]) and (not Q[13])),
        Q[15] == (Q[15] and (not Q[2])),
        Q[16] == (sum(Q[1:16]) >= 8),
        Q[17] == ((not Q[11] or not Q[15]) and not Q[19]),
        Q[18] == (Q[5] and Q[12]),
        Q[19], # Assume that multiple answers exist.
        Q[20] == (sum(Q[1:21]) == 10.5),
        Q[21] == ((not Q[9]) and Q[2] and (Q[1] or not Q[6]))
    ])

ANSWERS = []

# Adding Q[0] = None to avoid having to renumber all the subscripts
for Q in itertools.product(*([[None]] + [[False, True]] * 21)):
    if satisfied(Q):
        ANSWERS.append(Q[1:])

For which the true statements are:

Any of the following combinations: [2, 8, 9, 12, 19] [2, 6, 7, 9, 12, 13, 19] [2, 6, 7, 9, 11, 12, 13, 19] [2, 5, 8, 12, 18, 19, 21] [1, 4, 6, 9, 12, 19] [1, 4, 6, 9, 12, 15, 19] [1, 4, 6, 9, 11, 12, 19] [1, 4, 6, 9, 11, 12, 15, 19] [1, 2, 6, 9, 12, 19] [1, 2, 6, 9, 11, 12, 19]

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  • 1
    $\begingroup$ Q[1:20] will look at questions 1 to 19 inclusive. Surely that isn't what you want? $\endgroup$
    – Gareth McCaughan ♦
    Commented Aug 18, 2018 at 13:09
  • $\begingroup$ Similarly for Q[1:15]. $\endgroup$
    – Gareth McCaughan ♦
    Commented Aug 18, 2018 at 13:09
  • $\begingroup$ @GarethMcCaughan: You're right. I've edited my program accordingly. $\endgroup$
    – dan04
    Commented Aug 18, 2018 at 15:24
  • $\begingroup$ There are valid solutions with Q17=true. How come you haven't any? This is probably because "... and not q19" is rather "... and q19". BTW, my understanding of Q17 ("Either 11 or 15 is false") is that both 11 and 15 cannot be false for 17 to be true (q17 = q11 XOR q15). $\endgroup$
    – xhienne
    Commented Aug 18, 2018 at 21:50
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    $\begingroup$ In Python, a range like Q[1:15] has an inclusive left end and an exclusive right end: it means elements 1 to 14. (If this seems weird and counterintuitive, consider that e.g. it means that the length of a[i:j] is exactly j-i, and that a[i:k] is the same as a[i:j] + a[j:k].) $\endgroup$
    – Gareth McCaughan ♦
    Commented Aug 18, 2018 at 21:58
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First, 20 is false, so is then 3 and 12 is true.

The number of statements can be effectively reduced:

7 says 13 is true. So 7 and 13 have the same truth value, and we can just consider them as one. If another statement refers to 13, we change it to 7.

9 says 5 is false. So 5 and 9 are opposite, we can treat them as one (the 5); if another statement refers to 9, we change it to the opposite reference to 5.

6 and 8 are opposites, like with 5 and 9 we drop the 8.

Because 12 is true, 2 says 4 is false. They are opposite, dropping 2. Similarly, 18 says that 5 is true, so 18 is dropped.

14 says [...] and 12 is false. But 12 is true so 14 is false, so is 10, claiming that 14 is true.

So far: 3 F, 10 F, 12 T, 14 F, 20 F

The puzzle shrinks to 11 statements:

1: 7 F and 6 T
4: 1 T and 4 T
5: 5 T and 6 F
6: (says nothing)
7: 1 F and 6 T
11: 11 T and 6 T
15: 15 T and 4 F
16: (to be dealt with later)
17: (11 F or/xor 15 F) and 19 T (?)
19: (to be dealt with later)
21: 5 T and 4 F and (if 6 T then 1 T)

If 5 is true, then, as 6 is false, the (if) bit of 21 is also true. So 21 can be simplified into: 5 T and 4 F

From 5's claim we conclude: 5 T implies 6 F
From 11's claim we conclude: 11 T implies 6 T
So 5 and 11 can't both be true, as their implications clash. Also 5 and 6 can't both be true.
From 15's claim we conclude that 15 and 4 can't both be true.
From 1's claim we conclude that 1 and 7 can't both be true.

If 6 is false, then so are 1, 7, 11, 13, ...; that makes at least 8 of statements before 16 false. But if 6 is true, then 8 and 5 are false, so is 1 or 7, so is 2 or 4. 11 is undecided, so it's false as it's not 100% true; that again makes at least 8 false statements before 16. So 16 is false, there are more false than true.

Whatever 19 is, the following combinations (with 17 having the same value as 19):

1F 2T 3F 4F 5T 6F 7F 8T 9F 10F 11F 12T 13F 14F 15F 16F 17:[same as 19 if OR/false if XOR] 18T 20F 21F
1T 2T 3F 4F 5F 6T 7F 8F 9T 10F 11T 12T 13F 14F 15T 16F 17F 18F 20F 21T

all satisfy the conditions. That makes 19 true - more than 1 answer. Fixed ones (proven): 3F 10F 12T 14F 16F 19T 20F. Can all others vary, too? 2, 4, 7, 13 are the only ones we don't know yet. So here are 4 possible answers, showing that none but the already proven ones are fixed (with the first one '17?' depends on whether it's OR or XOR):

1F 2T 3F 4F 5T 6F 7F 8T 9F 10F 11F 12T 13F 14F 15F 16F 17? 18T 19T 20F 21F
1T 2T 3F 4F 5F 6T 7F 8F 9T 10F 11T 12T 13F 14F 15T 16F 17F 18F 19T 20F 21T
1T 2F 3F 4T 5F 6T 7F 8F 9T 10F 11T 12T 13F 14F 15F 16F 17T 18F 19T 20F 21F
1F 2T 3F 4F 5T 6T 7T 8F 9F 10F 11T 12T 13T 14F 15T 16F 17F 18T 19T 20F 21F

"If a statement is not 100% right, it's false." - Should I understand it as: the ones that are undecided as seen above, are false?

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