Question 1108009
<pre>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</pre>