Solution : First concentrating on the statements 3,5,7. If statement 3 is to be satisfied, then 5 and 7 cannot be satisfied. If we consider the statement 3 to be true then the digits of the number must form a fibonacci series : 0112358. But since there are no constraints on the length of the numbr and the last two digits are not squares we eliminate 8. Therfore we have 011235 now. 5 cannot be included because the third digit is not the difference of first and second. We have 01123. Since there are two adjacent one's, zero cannot be included. Thus we are left with 1123, which satisfies all other conditions.
Solution Submitted by Arun Saragadam, IIT Kharagpur :
(Appreciating the effort put in to check all the cases and arrive at a single solution)
Lets assume 4 is true.Then the possible cases are:->Last three digits are 222,333,555,777,235,237,357 etcBut all 333 or 444 or 555 or 777 cant exist as the rules described by them will not let them exist. Similarly no two of them can coexist as one rule contradicts the other.Even 222 cant occur or atleast the number cant end in 2 as the 2 states the number is not even. Hence the possibilities are 223,233 etcEven this is not possible as their rules are against these.Hence, 4 is eliminated.
Lets assume now 0 is true.Hence, no two adjacent digits are equal which denies the existence of 0 along with 3,5,7 as according to their rules atleast two digits must be adjacent if 0 is one of the digits.Hence, 3,5,7 are eliminatedlets assume 1 is presentbut it says atleast one odd prime must be present Hence 1 is also eliminatedLets assume 2 is present which says the number is odd which requires the presence of 1,3,5,7 which eliminates 2 too.Lets assume 8 is present which says last two numbers are squares. hence they must be 09,90(00 and 99 are not possible as adjacent numbers must not be equal). But if 9 is present last two numbers must be even. Hence no other possibility exists. Hence 8 is also elimiinated.
Lets assume 9 is present.which says last two numbers must be even but we are left with no possibilities which eliminates 9 too.As 0 cant exist independently, this case is eliminatedNow, 0 and 4 are eliminated and we are left with 1,2,3,4,5,6,7,8,9
Lets assume 8 is presentLast two numbers can be therefore 11, 19, 99,91But whenever 9 is present it requires last two numbers to be even which leaves us with the only possibility of 11But 1 demands the presence of at least one of 3,5,7Suppose 3 is present 11 must be preceeded by 0 in order to satisfy rule no 3 hence 3 is eliminatedSuppose 5 is present 11 must be preceeded by 2 in order to satisfy rule no 5 and 3 before 2 (in order to reach 8) but 3 and 5 cant coexist hence 5 is eliminated too.Similarly the case of 7 is eliminated.Hence 8 is eliminated.
So, now 0,4,8 are eliminated and we are left with 1,2,3,5,6,7,9
Lets assume 9 is present Hence the last two digits can be 2 and 6 but if 2 is present the number cant be even which contradicts the possibility of 9. Hence 9 is eliminated.
So, now 1,4,8,9 are eliminated and we are left with 1,2,3,5,6,7
Suppose 6 is presentThere must be only one digit which is prime(2,3,5,7)If its 2 , the number cant be even and demands an odd number which must be 1 as other prime cant exit but 1 itself demands atleast one odd prime number which thus eliminates the possibility of 2.Suppose 3 or 5 or 7 is presentthere must exist another number which is nothing but 1 (as otherwise if only two numbers exist it means all the three rules 3,5,7 are true as there will be only two numbers and demands all the three which contradicts the rules)So, if 1 is present6 must be at the end as the number must be even otherwise 2 needs to be presentand there must be 6,1 and one of 3,5,6 as according to rule 6 there must be only one prime numbersuppose 3 exists3 and 1 cant add upto 6 hence 3 is eliminated5 -1 != 6 hence 5 is eliminated7-1 =6 but atleast one number must be present twice in order to prevent 0 from appearing but which is impossible as to satisfy rule 7hence 6 is eliminatedNow, we are left with 1,2,3,5,7
suppose 5 is presentaccording to 5 difference of two adjacent digits gives the third digit but here atleast one digit is to be present twice which is impossible with satisfying rule 5similarly the case with 7
Hence, 5,7 are eliminated
So, now we are left with 1,2,3Suppose 2 is presentthere should be another odd number present as the number cant be even according to rule no 2So, it should be 1 or 3 either case demands the third numberso all the three numbers must be presenthence in order to satisfy rule no 3 only the following number can be formed1123Hence, the answer is 1123
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