Difference between revisions of "2023 AMC 10A Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | Multiples of 5 always end in 0 or 5 and since it is a three digit number, it cannot start with 0. All possibilities have to be in the range from 7 x 72 to 7 x 85 | + | Multiples of <math>5</math> always end in <math>0</math> or <math>5</math> and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. All possibilities have to be in the range from <math>7 x 72</math> to <math>7 x 85</math> inclusive. |
− | ~walmartbrian ~Shontai ~andliu766 | + | <math>85 - 72 + 1 = 14</math>. <math>\boxed{(B)}</math>. |
+ | |||
+ | ~walmartbrian ~Shontai ~andliu766 ~andyluo | ||
==Solution 2 (solution 1 but more thorough + alternate way)== | ==Solution 2 (solution 1 but more thorough + alternate way)== |
Revision as of 20:55, 9 November 2023
How many three-digit positive integers satisfy the following properties?
- The number
is divisible by
.
- The number formed by reversing the digits of
is divisble by
.
Solution 1
Multiples of always end in
or
and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. All possibilities have to be in the range from
to
inclusive.
.
.
~walmartbrian ~Shontai ~andliu766 ~andyluo
Solution 2 (solution 1 but more thorough + alternate way)
Let We know that
is divisible by
, so
is either
or
. However, since
is the first digit of the three-digit number
, it can not be
, so therefore,
. Thus,
There are no further restrictions on digits
and
aside from
being divisible by
.
The smallest possible is
. The next smallest
is
, then
, and so on, all the way up to
. Thus, our set of possible
is
. Dividing by
for each of the terms will not affect the cardinality of this set, so we do so and get
. We subtract
from each of the terms, again leaving the cardinality unchanged. We end up with
, which has a cardinality of
. Therefore, our answer is
Alternate solution:
We first proceed as in the above solution, up to .
We then use modular arithmetic:
We know that . We then look at each possible value of
:
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
If , then
must be
or
.
If , then
must be
.
Each of these cases are unique, so there are a total of