2023 AMC 10A Problems/Problem 12
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 .
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.
. .
(Add 1 to include 72)
~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 by Math-X (First fully understand the problem!!!)
https://youtu.be/GP-DYudh5qU?si=t4QMuoYyk2u5n64a&t=3140
~Math-X
Video Solution ⚡️ 2 min solution ⚡️
~Education, the Study of Everything
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
https://www.youtube.com/watch?v=4uKo5NR2o9Y
-paixiao
Video Solution
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 |
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