1. At least one of the last two statements on this list is true.
2. Either this is the first true statement, or it is the first false statement on the list.
3. This list contains at least three consecutive false statements.
4. The difference between the number of the last true statement and the
   number of the first true statement is a divisor of n.
5. The sum of the numbers of the true statements equals n.
6. This is not the last true statement.
7. The number of each of the true statements is a divisor of n.
8. Exactly n% of the statements on this list are true.
9. The number of divisors of n that are bigger than 1 and less than n is
   greater than the sum of the numbers of the true statements.
10. This list does not contain three consecutive true statements.
    

The problem for April:

Find n.

Solution to MP32

We received correct solutions from Martín Argerami (Regina), Pierre Bornsztein (France), Normand Laliberté (Ontario), Patrick J. LoPresti (Massachusetts), Juan Mir Pieras (Spain), and Alexander Potapenko (Russia). Our treatment combines ideas from all these correspondents; it unfolds in five steps. Before starting, we have to agree that any statement must be either true or false; more precisely, if a statement is not true, then it is necessarily false and, conversely, if a statement is not false, then it is necessarily true. We shall return to this matter later.

Step 1. Statement 1 is false.
If S2 were true, then it would be the first true statement, implying that S1 is false. If S2 were false, then it would not be the first false statement, so S1 would have to be false. Thus, no matter what the truth-value of S2 might be, S1 must be false. From knowing that S1 is false we immediately deduce that S9 and S10 are also false. In summary, here is what we know to this point.

Statement	TruthValue	Confirmed Implications
1	False	Both S9 and S10 are false.
9	False	The number of nontrivial divisors of n is at most the sum of the numbers of the true statements.
10	False	The list contains 3 consecutive true statements.

Step 2. Statement 6 is true.
S6 cannot be false: Were it false, then it would be the last true statement, which contradicts itself. Thus S6 must be true. Since from step 1 we know that statements 9 and 10 are false, we deduce here that

Step 3. S7 is true and S8 is false.
We easily see that exactly one of S7 or S8 is true: If S7 and S8 were both true, then from S7 (together with step 2) we get n ≥ 6·7·4> 100, which contradicts S8. From step 2 we know that at least one of them is true, hence we must decide between two possibilities:

Case A. S8 is true and S7 is false.
Case B. S7 is true and S8 is false.

First suppose that we are in case A; we quickly see that this situation is not compatible with S3. Were S3 true (while S8 is true and S7 false) we would have

With this arrangement there would be no room for three consecutive false statements, which contradicts S3 (that there are at least three consecutive false statements).
On the other hand, were S3 false (while S8 is true and S7 false),

With S3 false there could be no more than two consecutive false statements, so S2 would necessarily be true, and (from the falsity of S10 in step 1) S4 and S5 would be true so there could be three consecutive true statements. In this hypothetical situation the sum of the true statements would be 2+4+5+6+8 = 25, contradicting S8 (which says that n = 50% of the statements are true). Conclusion: We cannot have Case A.

Therefore, S7 is true and S8 false. With S8 false, S3 becomes true (the list in fact contains three consecutive false statements). Our current state of knowledge is therefore,

Step 4. S5 is false.
Because S6 and S7 are true (steps 2 and 3), we now know that n is a multiple of 6·7 = 42, so n ≥ 42. S5 says that n is the sum of the statement numbers of the true statements, but the largest that sum could be (at this point) is 2+3+4+5+6+7 = 27 < 42; therefore, S5 is false. Moreover, since we know S10 is false from step 1, we must have three consecutive true statements so that S2 and S4 must be true. Consequently, we now know the truth-value of all ten statements:

Step 5. Compute n.
Here is a summary of the known facts about n.

Statement	TruthValue	Confirmed Implications
4	True	The difference between the numbers of the last true statement and the first true statement is a divisor of n: 7 – 2 = 5 divides n.
7	True	The number of each of the true statements divides n: 2, 3, 4, 6, and 7 all divide n, so (together with 5) n = 22·3·5·7·k = 420k for some k ≥ 1.
9	False	The number of nontrivial divisors of n is at most the sum of the numbers of the true statements.

To complete the solution, we now use the negation of S9. 420 has 24 divisors, so it has 22 nontrivial divisors. That is, n = 420k has at least 22 nontrivial divisors. The sum of the numbers of the true statements is 2+3+4+6+7 = 22which (by S9 false) can be no smaller than the number of nontrivial divisors of n. Thus, 420 already achieves the maximum number of 22 divisors. We conclude that k = 1 and therefore, n = 420.

Further comments.
We learned of the problem from a Belgian colleague who misfiled the reference after having translated it from the English original when it appeared back in the 1980s. We have no way of finding out who devised this marvelous problem; perhaps someone reading this page can tell us the author and where the problem was originally published.

LoPresti raised two issues. First he pointed out that zero is also a solution -- zero is certainly a number, and it satisfies all the conditions implied by the ten statements. (The statement of our problem on the French page stated explicitly that n must be positive, which immediately eliminates zero.) Although one generally assumes that the word number refers to a positive integer in any context involving counting and divisibility, this is mathematics and we should avoid ambiguity as much as is practical. We should have stated that n is a positive integer (as was intended).

The other point he raised is deeper and more serious. He wonders whether self-referential statements can be assumed to have a well-defined truth-value. Surely a sentence such as,

"This statement is false."

cannot be permitted -- if the statement were true then, by what it says, it must be false, while if the statement were false then, by what it says, it must be true. On the other hand, there is no need to ban all self-referential statements. One delightful feature of our system of ten interconnected statements is that there does exist a way to consistently assign truth-values to each of the statements. Furthermore, as our solution shows, that assignment of truth-values is the only assignment that is consistent. The problems caused by self-referential statements helped, at the turn of the last century, to motivate a re-examination of the foundations of mathematics; today, a century later, there still is no completely satisfactory resolution of these difficult problems. The good news is that most of mathematics is far removed from these difficulties, and we can blissfully do mathematics, leaving it to our logicians to worry about the foundations.