Difference between revisions of "2023 AMC 10A Problems/Problem 12"
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~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS | ~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS |
Revision as of 11:35, 11 November 2023
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 .
~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.
. .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.