The above solution is incorrect because it ignores the restriction that a, b, and c are DIFFERENT one-digit whole numbers. Here is the correct solution: Note: I will use the rule followed in " 9abc2 ". That is, if no multiplication symbol is placed between either two letters or between a digit and a letter, then the letter is assumed to be a digit and no multiplication symbol is ever understood and not written. Multiplication will only be assumed if the symbol "×" appears. 9abc2 = 90002 + abc0 Since 90002 is divisible by 11, [90002 = 11×8182] 9abc2 will be divisible by 11 if and only if abc0 is also divisible by 11. Since abc0 = 10×abc, abc0 will be divisible by 11 if and only if abc is divisible by 11. Therefore if abc is any term of this arithmetic sequence S = 000, 011,022,...,099,110,121,132,...,968,979,990, then 9abc2 will be divisible by 11 If we divide every term of sequence S by 11, we get the arithmetic sequence: 0,1,2,...,9,10,11,12,...,88,89,90 That's 91 terms, since it starts with 0 and ends with 90. So if we did not have the requirement that a,b, and c must all three be different digits, the answer would be 91 [Note: this 91 is what the other tutor gave as a final solution.] But from the 91 terms of multiples of 11 (less than 1000), we must eliminate all that have 2 or three digits the same: Case 1: a=b, that is abc = aac then abc = 100×a+10×a+c abc = 110×a + c abc-110×a = c Both terms on the left are divisible by 11, so c must be a digit divisible by 11, and 0 is the only digit divisible by 11. Thus c=0, So we must remove this arithmetic sequence: 000,110,220,...,990 If we divide every term of that arithmetic sequence by 11, we get the arithmetic sequence 0,1,2,...,9 That's 10 terms, since it starts with 0 and ends with 9. Thus by case 1, we remove 10 terms from sequence S. ------- Case 2: b=c, that is abc = abb then abc = 100×a+10×b+b abc = 100×a + 11×b Since abc is divisible by 11, there exists integer K such that abc = 11×K 11×K = 100×a + 11×b 11×K - 11×b = 100×a Both terms on the left are divisible by 11, and 100 is not divisible by 11, so a must be a digit divisible by 11, and 0 is the only digit divisible by 11. Thus a=0, So we must remove this arithmetic sequence: 000,011,022,...,099 We have already removed 000 in case 1, so in case 2 we need remove only the other 9 terms. Thus for case 2, we remove 9 terms from sequence S. Case 3: a=c, that is abc = aba [The most difficult case]. then abc = 100×a+10×b+a abc = 101×a + 10×b Since abc is divisible by 11, there exists integer K such that abc = 11×K 11×K = 101×a + 10×b We write 101×a as 121×a - 22×a +2×a, and 10×b as 11×b-b 11×K = 121×a - 22×a +2×a + 11×b-b We divide through by 11 K = 11×a - 2×a + (2/11)×a + b - c/11 We isolate the fractions on the left: c/11 - (2/11)×a = 11×a - 2×a + b - K The right side is an integer, so the left side must also be an integer, say the integer P c/11 - (2/11)×a = P Multiplying through by 11 c - 2×a = 11×P c = 2×a + 11×P <-- this could also be written c = (2×a) mod 11 P could only be 0 or -1, for no other values will allow c to be a digit. Sub-case 3A: P = 0 c = 2×a, this gives only possibilities a=0,1,2,3,4 and we have already counted 000 in case 1. Thus for sub-case 3A, we remove 4 more terms from sequence S. [FYI, the 4 terms removed from S here are 121, 242, 363, 484] Sub-case 3B: P = -1 c = 2×a-11, this gives only possibilities a=6,7,8,9. Thus for sub-case 3B, we remove 4 more terms from sequence S. [FYI, the 4 terms removed from S here are 616, 737, 858, 979] Thus for case 3, we remove 4+4=8 terms from sequence S. So the final answer is 91-10-9-8 = 64. Edwin