Also of true and false statements

(first solution) Let S1 denote statement 1 , S2 denote statement 2 and so on and also let the truth values of this statements be noted with T for true and F for false. Also for enhancing clarity sometimes I will write what each statement says when I speak of it to see what is deduced and how and I hope it is clear.

For enhancing clarity as a fast synopsis what will be done is the following. Firstly deduce that whatever value S2 has it implies 1F which implies that 9 and 10 are false. Then also by the auto-referential propriety deduce that S6 is true. Check S3 , which leads to study of 2 cases. Case 1 leads to N = 420 and case 2 leads to an inconsistent set of truth values seen by what the statements mean. Therefore case 1 is good and therefore the only result.

Start the reasoning from S2 which says that "this statement is either the first true statement or the first false statement" and observe that however this statement is it is necessary that S1 is false. If statement 2 is true then it is "the first true statement" and therefore all statements before it are false , meaning that statement 1 must be false (1) and if statement 2 is false then it is not true that it is "the first false statement" meaning that at least one statement before it is false and therefore S1 is false (2). So , no matter how S2 is S1 is false , therefore S1 is necessarily false. Because S1 is false then none of S9 and S10 are true therefore S9 and S10 are false and then "the number of divisors of N is smaller than the sum of the numbers of true statements" (A1) and "there are at least three consecutive statements" (A2).

So until now it is known that S1 is false and because of this S9 and S10 are false.

Observe that S6 ("this is not the last true statement") must be true because otherwise S6 is the last true statement which will be a contradiction. Therefore S6 is true and therefore because , being true , this statement is not the last true statement at least one of S10 , S9 , S8 or S7 must be true. S10 and S9 can't be true because it is known that they are false , therefore at least one of S8 and S7 are true. So up to this point it was deduced that S10 , S9 S1 are false and S6 is true with at least one of S8 and S7 true.

To deduce the rest of the truth values of the other statements consider S3 which speaks about the way this values look like because it states that "there are 3 consecutive false statements".

(Case 1) Suppose S3 is true then what it says is true namely that there are 3 consecutive false statements. It is known that 10F 9F 8 7 6T 5 4 3T 2 1F and there must be 3 consecutive false statements. From this and from the fact that at least one of S8 and S7 is true the only possible configuration for 3 consecutive false statements is 10F 9F 8F while 7T because 8 must be false in order to have 3 consecutive false statements but 7 must be true to respect the conclusion from S6 T.

So , until now it is known that 10F 9F 8F 7T 6T 5 4 3T 2 1F. Now , supposing 5 is true from 7 as a true statement would result that N divides 3 , 5 , 7 namely 3 * 5 * 7 =105 while from 5 would result that N being the "sum of the true statements" N < 55 (because at most all the 10 statements added will give 10! =55) and therefore N should be less than 55 but divide 105 which is not possible. From this , 5 can't be true and therefore it is false and therefore it has been deduced in this step that 5F. Filling the table it can be therefore presented the situation as 10F 9F 8F 7T 6T 5F 4 3T 2 1F.

Then from A2 the remaining values to be filled in the table S4 and S2 are true. Then , from S4 the number divides also with 7-2=5 and combined with S7 divides all true statements therefore N divides the numbers associated to the true statements 7 , 6 , 5 , 4 , 3 and 2 and therefore divides 420 meaning that the number looked for is a multiple of 420 or N =420m.

Finally the table with all results which cover all the information needed to deduce the number providing that the problem is correct can be written and is 10F 9F 8F 7T 6T 5F 4T 3T 2T 1F.

Because the sum of the true statements is 22 from A1 N should have a number of divisors either equal or less than 22 and because 420 = 2 ^ 3 *5 * 7 it has (2+1) * (1+1) * (1+1) * (1+1) - 2 = 22 divisors which checks A1. Since 420 has 22 divisors a number of the form 420m would have more and therefore 420 is the only case which checks all conditions of the problem for the consistent set of truth values and statements.

(Case 2 , that is) Suppose 3 is false. The table will look at first like 10F9F8 7 6T 5 4 3F 2 1F and also there aren't going to be 3 consecutive false statements meaning that there are at most 2 false statements and then because 2 and 8 if false will produce a configuration of 3 false statements (look at the table) must be true and at least one of 4 and 5 is true because otherwise there will be again 3 false consecutive statements. For enhancing clarity to remind S2 and S8 T and at least one of S4 and S5 are true.

Because S8 is true then N < 100.

For 7 true the values will look like 10F 9F 8T 7T 6T 5 4 3T 2T 1F and then N divides at least 8 , 7 , 6 and 2 then is a multiple of 8 * 7 * 6 * 2 = 168 which is bigger than 100. This contradicts S8 so S7 must be false. For 7 false the configuration will be 10F 9F 8T 7F 6T 5 4 3T 2T 1F and S5 and S4 because of statement A2 must be true meaning by S5 true than N = 2+4+5+6+8 and from 4 that N divides 8-2=6. This is a contradiction as 25 doesn't divide 6 (of course) anyways.

Then for S8 true however S7 is the configuration leads to a contradiction because the statements for the assigned truth values are not consistent with each other. Then for case 2 a configuration is impossible and therefore just for the first case it is possible.

Since from Case 1 and Case 2 everything has been verified as a function for the value of S3 it can be concluded that just one solution , namely N=420 in case 1 works. Therefore it is possible for N=420.

The number N is therefore 420 and the answer to the problem is 420 anyways.

(Second solution) As this problem asks for identifying a number which is determined by knowing something about the true and false statements it must be that firstly there should be identified at least some truth values of statements. This is implicitly considered into the question as the right approach but it is good make explicit such that it is clear what is being done completely.

The solution of this problem starts by firstly assigning necessary truth values to some statements such that there is a point from which to guide the rest of reasoning. The 2 statements to which it is better to look firstly tend to be the 2 auto-referential statements 2 and statement 6.

Firstly analyze statement 2 which can be either true or false. For the first case observe that if statement 2 is true then it means that what it states by it's auto-referentiality namely that is "either the first true statement or the first false statement" is true and as it is true that it is the first true statement it will imply that statement 1(which is the only one before it) must be not true and therefore statement 1 must be false and for the other case which should be analyzed in which statement 2 is false observe that also by it's own auto-referentiality as it will not be true that it is "the first false statement" it must be that there is another statement before it which is false which being just one statement (namely 1) before it implies statement 1 is false leading to the the conclusion that however statement 2 is statement 1 must be false and to articulate therefore being necessary that statement 1 is false. If statement 1 is false then statements 9 and 10 are false meaning that the number of divisors of N is less or equal than the sum of the numbers associated to the true statements (A1) this being the contradiction of 9 and there are 3 true consecutive statements this being the contradiction of statement 10(A2) this 2 contradictions being noted with A1 and A2 for the purposes of the presentation as they are going to be used multiple times along the reasoning that will lead to the solution of the problem.

After identifying that statements 1 , 9 , 10 must necessarily be false analyze the other auto-referential statement , namely the statement 6. Observe that statement 6 is certainly true as if it would be a false statement 6 is the last true statement which leads to a contradiction and therefore 6 what it is stated namely that it is not the last true statement is true meaning that after it will be at least one other statement which is true and as 9 and 10 are already false it must be that at least one of 8 or 7 are true.

Until now , what is known is the following situation 1F 6T 9F 10F. To look for the other values of the rest of the statements and identify which are true and which are false consider statement 3. As it does speak about the configuration of the true and false statements it will provide some guiding around it. What this means is that it will make easier to find the truth values because gives some idea about them anyways.

If statement 3 will be true then there must be 3 consecutive false statements and since statements 3 and 6 (being just 2 statements before any of statements 3 and 6) are true the only possible configurations for 3 consecutive false statements will be that of either the sequence 8 9 10 or 8 9 10 and 7 are false but since one of statements 8 and 7 are true the only sequence remains 8 9 and 10 false and 7 that will be true. If 5 would be true then from statement 7 which is true would be concluded that N divides 3 * 5 * 7 = 105 and by statement 5 that N will be a number which is the sum of all true statements which can't be greater that 55 ( the sum of all numbers until 10 being 10!=55) which will therefore not be a multiples of 105 so 5 is false. Then , as there are 3 true consecutive statements (A1) and 5 is proved to be false the configuration until now looking with statements 8 , 9 , 10 , 5 false and 6 , 7 true the only possible configuration of 3 consecutive true statements remaining is that of 2 , 4 , 3 which should be therefore all true. As 2 and 4 are true N will divide among other divisor also with 7-2 = 5 (from 4) and will divide 2 , 4 , 3 , 5 , 6 , 7 being a multiple of the bcmm 420. This completes the configuration as all truth values of the statements have been completed where the sum of the numbers 2 + 4 + 3 + 7 = 22 and as statement 9 is false the number N should be a number which is a multiple of 22 but has less or an equal number of divisors with 22 meaning that since 420 already has 22 divisors a number of the form 420 * a will have at least one more (a) and there is only one solution.

To check if this is the only solution consider the case in which statement 3 is false in which will not be a the 3 false statements meaning that then 2 and 8 will be true and also at least one of 4 and 5 will be true. Considering the consequences of this observe that since 8 is true then the true statements will be N% from the the 10 statements and therefore N < 100 and taking it in conjunction with the statement number 7 if it is true then the number N divides 2 6 8 and 7 thing which would imply that the number N is a multiple of 168 which contradicts that N <100 therefore leading to an impossible case. For the other part for 8 true and for statement 7 being a false one then as there are 3 consecutive true statements (A2) statements in the configuration somewhere this will be possible just for the configuration of 4 5 and 6 meaning from 5 that number N = 2+4+5+6+8 which is 25 but from statement 4 it means that N divides 8-2=6 which will lead again to a contradiction since 25 doesn't divide 6 therefore being concluded that it will be indeed impossible for 3F as will imply 8 T and by conjunction with 7 whichever values has leads to a contradiction anyways.

Therefore the only case that works from the analysis done is for 3 being true because that is the only case in which there is no contradiction and has the consequences of consistent statements and solution.

This means that there is therefore only one possible situation which doesn't get to a contradiction or in which the set of all statements are consistent that being resulted in N=420 anyways.