Difference between revisions of "2023 AMC 10A Problems/Problem 12"
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We know that <math>0\le a,b<10</math>. We then look at each possible value of <math>a</math>: | We know that <math>0\le a,b<10</math>. We then look at each possible value of <math>a</math>: |
Revision as of 19:38, 9 November 2023
How many three-digit positive integers satisfy the following properties?
- The number is divisible by .
- The number formed by reversing the digits of is divisble by .
Solution 1
Multiples of 5 always end in 0 or 5 and since it is a three digit number, it cannot start with 0. All possibilities have to be in the range from 7 x 72 to 7 x 85 inclusive. 85 - 72 + 1 = 14. .
~walmartbrian ~Shontai ~andliu766
Solution 2 (solution 1 but more thorough + alternate way)
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
Alternate solution:
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