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>. | ||
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+ | ~A_MatheMagician | ||
==Solution 1== | ==Solution 1== |
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.
. .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.