2023 AMC 10A Problems/Problem 12
Contents
[hide]Problem
How many three-digit positive integers satisfy the following properties?
- The number
is divisible by
.
- The number formed by reversing the digits of
is divisible by
.
Note
One thing to note is the number 560. When it is flipped, the result is 065, which is a number but has a leading zero. Since the problem doesn't say anything about 560, it is assumed to be a valid . HamstPan38825 provides a good explanation on why this problem is wrong, "Define
to be the digit-reversal function in question, and suppose for the sake of contradiction that
is a strictly defined number, hence
, as was assumed when
was included in the count. Thus
and
are equivalent under input to
too, so
which is a contradiction as
is a function. Hence
is not a strictly-defined number, and it cannot be divisible by
."
~A_MatheMagician ~ESAOPS
Solution 1
Multiples of will always end in
or
, and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with
. Since the numbers must be divisible by 7, all possibilities have to be in the range from
to
inclusive.
.
.
~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS
Solution 2 (solution 1 but more thorough)
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
~ Technodoggo
Solution 3 (modular arithmetic)
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
~ Technodoggo
Solution 4
The key point is that when reversed, the number must start with a or a
based on the second restriction. But numbers can't start with a
.
So the problem is simply counting the number of multiples of in the
s.
, so the first multiple is
.
, so the last multiple is
.
Now, we just have to count .
We have a set that numbers 85-71 =
~Dilip ~boppitybop ~ESAOPS (LaTeX)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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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