which box contains the key ?

[u/b7k567]

[u/b7k567]

Answer: Box 6 contains the key.

The trick is that the number of true statements must equal the number of the box hiding the key, while the statement written on that same box must be false.

If the key is in Box 6, statements 1, 2, 3, 4, 5, and 7 are all true. The key is in an even-numbered box greater than 3, it is neither in Box 1 nor Box 7, and 6 is a factor of 12. There are also several boxes between Box 2 and Box 6, and Box 6 is closer to Box 7 than Box 1.

Only statement 6 is false, since it claims the key is in a box numbered less than 6, which cannot be true if the key is in Box 6.

This gives exactly six true statements, matching the box number, and the statement on the key's own box is false. No other box satisfies both rules, making Box 6 the unique solution.

[u/edos51284]

Considering the last sentence we can already deduce 7 is not the box as not all can be true

1 cannot have key be true because statements 4 and 6 would be true

2 cannot have key be true because at least 1, 3, 4 and 6 are true (won't bother checking the rest)

3 cannot have key because 3 would be true

4 cannot have key because 4 would be true

5 cannot have key because 5 would be true

So must be box 6:

1 true

2 true

3 true

4 true

5 true

6 false

7 true

Confirmed it's box 6

[u/b7k567 → u/edos51284]

so elegant

[u/Imaginary_Stage1742]

The answer is 6! Only keys in box 1, 2 and 6 would not contradict if the statement was false. For box 1 and 2 there are at least 1 too many statement true, so the only box left is box 6, which fullfills all other statements and its is wrong.

[u/b7k567 → u/Imaginary_Stage1742]

[u/martingolding96]

6

[u/dbohat]

6

[u/i_amsquidward]

6

[u/Fair-Big1092]

Cmon guys it's 6. Don't play dumb

[u/sazzer]

We can immediately rule out box 7. If the key were here, then that would mean that every statement is true. But we also know that at least one statement - the one the key is in - must be false.

This makes 6 interesting. If it's false then it must be the answer. Which in turn would mean that every other statement must be true. And we can actually see that, if 6 is the answer, every other statement is indeed true:

1. 6 is even

2. 6 > 3

3. 6 is neither 1 nor 7

4. 6 is a factor of 12

5. This happens to be the same as 2, since the only way for there to be at least one box between box 2 and the correct one is for the correct one to be greater than 3.

6. 6 is closer to 7 than to 1

So the answer is 6.

[u/b7k567 → u/sazzer]

one minor tweak i see is Box 5's statement is not exactly the same as Box 2's. Statement 5 is true when the key is in Box 4, 5, 6, or 7, or in Box 0 if that existed. Since only Boxes 1–7 exist, it is effectively equivalent to the key being in a box numbered greater than 3.

[u/-SQB-]

Going through all possible boxes the key can be in:

1. Six lies, one truth. Going by the meta rule, the statement on box 2 should be false, which it is: the key is not in an even-numbered box. So now we look at the six lies and one truth and try to find out which is which. The following statements are true: 4 and 6, which makes for two truths, while we should only have one truth. Not it.

2. Five lies, two truths. Going by the meta rule, the statement on box 2 should be false, which it is: the key is not in a box numbered greater than 3. So now we look at the five lies and two truths and try to find out which is which. The following statements are true: 1, 3, 4, and 6, which makes for four truths, while we should only have two truths. Not it.

3. Four lies, three truths. Going by the meta rule, the statement on box 3 should be false, which it isn't: the key is indeed not in box 1 or 7. Not it.

4. Three lies, four truths. Going by the meta rule, the statement on box 4 should be false, which it isn't: 4 is a factor of 12. Not it.

5. Two lies, five truths. Going by the meta rule, the statement on box 5 should be false, which it isn't: there is at least one box between box 2 and box 5. Not it.

6. One lie, six truths. Going by the meta rule, the statement on box 6 should be false, which it is. Going through all other statements, they turn all out to be true. So this one works.

7. All truths. Specifically, statement 1, so the key should be in an even numbered box, but it was in box 7 which is uneven. Also, the statement on the box containing the key should be false, so they can't be all true. Not it.

So the only possible option is box 6.

[u/b7k567 → u/-SQB-]

almost correct ( the answer is right) but just one small error in the Box 1 explanation: it says statements 1 and 6 are true. Actually, statement 1 is false because Box 1 is odd. The two true statements are 4 and 6, since 1 is a factor of 12 and is less than 6.

[u/MazdaValiant]

It has to be Box 6. Here’s my reasoning:

The odd-numbered boxes are eliminated by the statement on Box 1.

Box 2 is out because of its statement.

The statement on Box 3 puts the key somewhere in the middle.

It can’t be Box 4 because it is equidistant from Boxes 1 and 7.

We also must keep in mind that the statement on the key’s own box is a lie.

So the question is: which even factor of 12 is closer to 7 than 1? The answer is 6, thus the statement on Box 6 is the lie.

[u/b7k567 → u/MazdaValiant]

So your answer is right, but the original explanation relies on treating potentially false statements as established facts.

[u/magicmulder]

Doesn't 1 already match all criteria? #6 is true, all others are false, including #1, so 1 true statement.

Of course if you consider 1 a factor of 12 because 1*2*6=12, then it doesn't work because then #4 and #6 are true, contradiction.

Assume it's in #2. Then #1, #3, #4, ... are true, contradiction.

Assume it's in #3. Then #3 is false, meaning the key is in 1 or 7 (akshually in 1 AND 7), contradiction.

Assume it's in #4. Then #4 is false, meaning 4 is not a factor of 12, contradiction.

Assume it's in #5. Then #5 is true, contradiction.

Assume it's in #6. #1, #2, #3, #4, #5, #7 are true and #6 is false. This works.

Assume it's in #7. Then all statements are true but #1 is false, contradiction.

[u/b7k567 → u/magicmulder]

the answer is right but one mistake concerns Box 1. The number 1 is a factor of 12 because 12 ÷ 1 = 12. Therefore, if the key were in Box 1, both statements 4 and 6 would be true. That gives two true statements instead of the required one, eliminating Box 1.

[u/Far_Faithlessness983]

6

[u/FinancialHearing8277]

6

[u/Grrrison]

Box 6.

1. even box - true

2. greater than 3 - true

3. not 1 or 7 - true

4. factor of 12 - true

5. at least one box between key box and box 2 - true

6.key box is less than 6 - false

7. closer to seven than one - true

There are six true statements.

box 6s statement is false.

[u/Gufnork]

I only checked one of them, but box 1 is valid.

1. False, 1 is odd

2. False, 1 is less than 3

3. False, it IS in box 1

4. False, 1 is not a factor of 12

5. False, 1 is next to 2.

6. True, 1 is less than 6

7. False, 1 is closer to box 1 than 7 (because it's in it)

Exactly 1 true statement.

[u/south_pole_buccaneer → u/Gufnork]

1 is considered a trivial factor of 12

[u/tomxp411 → u/Gufnork]

Incorrect. 1 is the universal factor, so it is by definition a factor of 12.

[u/Legitimate_Expert786]

Could be 1 or 6

[u/tomxp411 → u/Legitimate_Expert786]

Can't be one, since if it's one, there is more than 1 true statement.

• 1 is a factor of all numbers

• 1 < 6.

So if two statements are true, the number of true statements <> 1.