Question 1108009: The five-digit number 9abc2 is divisible by 11, where a,b,and c are different one-digit whole numbers. How many combinations of digits are possible values of a,b,and c?
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
My first inclination, whenever I see a problem involving divisibility by 11, is to use the rule for divisibility by 11. That rule, applied to this problem, would say that either
(1) (9+b+2)-(a+c) = 0, or
(2) (9+b+2)-(a+c) = 11, or
(3) (9+b+2)-(a+c) = -11
Logical analysis shows there are 55 solutions of type (1) and 36 of type (2) for a total of 91; there are no solutions of type (3).
Completing the logical analysis to reach the answer by that method is good mental exercise.
However, for this particular problem, there is a much faster path to the answer.
The first and last digits are fixed; and the number 90002 is divisible by 11.
Since the last digit is fixed, to get from one number divisible by 11 to the next, you have to add 110.
So the answer to the problem is the number of times you can add 110 to 90002 and still have 9 as the leading digit.
 (approximately)
That calculation shows there are 90 more numbers after 90002 with leading digit 9 and last digit 2 that are divisible by 11; so the total number of numbers with those characteristics is 91.
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website! 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
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