Difference between revisions of "2023 AMC 10A Problems/Problem 12"
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<math>\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17</math> | <math>\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17</math> | ||
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+ | ==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 <math>N</math>. | ||
+ | |||
+ | ~A_MatheMagician | ||
==Solution 1== | ==Solution 1== |
Revision as of 12:35, 11 November 2023
Contents
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 .
~A_MatheMagician
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.
.
.
One thing to note is the number 560. When it is flipped, you get 065, and no one actually writes it like this. This problem doesn't specify whether or not 560 is allowed.
~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
Initially, I thought of finding that there are such numbers divisible by
since
divided by
gives
with a remainder. But it's not relevant!
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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