2017 AMC 10A Problems/Problem 25
Let the three-digit number be :
If a number is divisible by 11, then the difference between the sums of alternating digits is a multiple of 11.
There are two cases:
We now proceed to break down the cases.
:
. This has
cases.
:
, this case results in 110, 220, 330...990. There are two ways to arrange the digits in each of those numbers.
:
, this case results in 121, 231,... 891. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
, this case results in 242, 352,... 792. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
, this case results in 363, 473,...693. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
, this case results in 484 and 594. There are
ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to
cases.
:
.
:
, this cases results in 209, 308, ...506. There are
ways to arrange each of those cases. This leads to
cases.
:
, this cases results in 319, 418, ...616. There are
ways to arrange each of those cases, except the last. This leads to
cases.
:
, this cases results in 429, 528, ...617. There are
ways to arrange each of those cases. This leads to
cases.
...
If you continue this counting, you receive cases.