Difference between revisions of "2017 AMC 10A Problems/Problem 25"
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Let the three-digit number be <math>ACB</math>: | Let the three-digit number be <math>ACB</math>: | ||
− | If a number is divisible by 11, then the difference between the sums of alternating digits is a multiple of 11. | + | If a number is divisible by <math>11</math>, then the difference between the sums of alternating digits is a multiple of <math>11</math>. |
There are two cases: | There are two cases: |
Revision as of 17:47, 8 February 2017
Problem
How many integers between and
, inclusive, have the property that some permutation of its digits is a multiple of
between
and
For example, both
and
have this property.
Solution
Let the three-digit number be :
If a number is divisible by , then the difference between the sums of alternating digits is a multiple of
.
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.
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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