Difference between revisions of "2023 AMC 10A Problems/Problem 12"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
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+ | https://www.youtube.com/watch?v=7qWMFzq7Tfs | ||
==Video Solution== | ==Video Solution== |
Revision as of 06:59, 28 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (solution 1 but more thorough)
- 4 Solution 3 (modular arithmetic)
- 5 Solution 4
- 6 Video Solution
- 7 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 8 Video Solution
- 9 Video Solution by Math-X (First understand the problem!!!)
- 10 Video Solution by Power Solve (easy to digest!)
- 11 Note
- 12 See Also
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
.
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
~Dilip ~boppitybop ~ESAOPS (LaTeX)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=7qWMFzq7Tfs
Video Solution
https://www.youtube.com/watch?v=4uKo5NR2o9Y
-paixiao
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/N2lyYRMuZuk?si=6B-mTB070UP2yuDF&t=435
~Math-X
Video Solution by Power Solve (easy to digest!)
https://www.youtube.com/watch?v=Mg6JUanYNJY
Note
According to the official answer key, choice (B) is correct. However, some have argued that it is ambiguous whether the number should be included in the count, since its reversal,
, has a leading zero. It is assumed that
denotes the two-digit number
, which is divisible by
, but MAA should have clarified what happens when a number with trailing zeros is reversed.
~A_MatheMagician ~ESAOPS ~sdpandit
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.