2023 AMC 10A Problems/Problem 12
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 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 inclusive. 85 - 72 + 1 = 14. .
~walmartbrian ~Shontai ~andliu766
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:
\begin{align*} 0&\equiv N~(\mod7) \\ &\equiv500+10a+b~(\mod7) \\ &\equiv3+3a+b~(\mod7) \\ 3a+b&\equiv-3~(\mod7) \\ &\equiv4~(\mod7). \\ \end{align*}
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