Riddles, brain puzzles and mathematical problems

[MyRedNeptune]

@MyRedNeptune
Just the answer won't do it. You have to describe how you found it.

Sorry, was in a hurry so didn't have time to type it up.

I noticed that you have edited the text of sentence 4. I will adjust my answer accordingly.


First, let's look at sentence 2. If it is true, then it must be the first true sentence, therefore sentence 1 must be false. If it is false, then it cannot be the first false sentence. Therefore, sentence 1 must be false. It follows, then, that sentence 1 is false at all times.

If sentence 1 is false, then both sentences 9 and 10 are false.

Sentence 6 can only be true. If it were false, then it would be not the last true sentence, and therefore true, which is a paradox.


What we have now is this initial setup:

1. f
2.
3.
4.
5.
6. t
7.
8.
9. f
10. f


In other words:

-there are 3 successive true sentences
-the number of positive (and, I'm assuming, natural) divisors of w, excluding 1 and w, is greater than the sum of the numbers of all true sentences
-there is at least one true sentence after sentence 6


With this in mind, we can move on to the different cases.

Case I.

Spoiler below!

Let's assume that 2 is false.

a) If 2 is false then 3 cannot be false and is therefore true.
b) Because 3 is true, 8 must be false or it is impossible to satisfy the condition.
c) Because 6 is true and 8, 9 and 10 are false, then 7 is true.

Now we have this:

1. f
2. f
3. t
4.
5.
6. t
7. t
8. f
9. f
10. f

d) Because 10 is false, then 5 must be true. If 5 is true, then w is either 21 or 25, neither of which satisfy the condition of sentence 7.

Therefore, 2 cannot be false.

Case II.

Spoiler below!

Since we have discovered that 2 cannot be false, we now consider it to be true. Let's assume that 3 is false.

a) If 3 is false then 8 must be true. This means that w is a multiple of 10.

1. f
2. t
3. f
4.
5.
6. t
7.
8. t
9. f
10. f

b) If 3 is false then at least one out of 4 and 5 must be true.
c) If 4 is true, then w must be a multiple of 6 and 10 - the only possible numbers being 30 or 60. However, if 4 is true, then w > 30. If w = 60, then 5 and 7 must also be true. 5 and 7 cannot be true when w = 60.
d) If 5 and not 4 is true, then w is either 21 or 28, which are not multiples of 10.

Therefore, 3 cannot be false.

Case III.

Spoiler below!

2 and 3 are true.

a) Because 3 is true, 8 must be false.
b) Because 8 is false and 6 is true, 7 must be true.

1. f
2. t
3. t
4.
5.
6. t
7. t
8. f
9. f
10. f

c) because 10 is false, at least one out of 4 and 5 must be true.
d) If 5 is true, then w is either 23 or 27, which does not satisfy the condition of sentence 7. Therefore, 5 is false and 4 is true.

5*7*4*3 = 420

The number 420 has 22 divisors (2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210) and the sum of the numbers of all true sentences is 22. This is consistent with sentence 9.

Therefore, w = 420

Also,

---

(I hope I haven't made any stupid mistake here)

9c = 9b + 9 <=>
c = b + 9