Difference between revisions of "2024 AMC 10A Problems/Problem 18"
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We now simply must count the number of integers between <math>5</math> and <math>2024</math>, inclusive, that are <math>6</math> mod <math>8</math> or <math>3</math> mod <math>4</math>. Note that the former case comprises even numbers only while the latter is only odd; thus, there is no overlap and we can safely count the number of each and add them. | We now simply must count the number of integers between <math>5</math> and <math>2024</math>, inclusive, that are <math>6</math> mod <math>8</math> or <math>3</math> mod <math>4</math>. Note that the former case comprises even numbers only while the latter is only odd; thus, there is no overlap and we can safely count the number of each and add them. | ||
− | In the former case, we have the numbers <math>6,14,22,30,\dots,2022</math>; this list is equivalent to <math>8,16,24,32,\dots,2024\cong1, | + | In the former case, we have the numbers <math>6,14,22,30,\dots,2022</math>; this list is equivalent to <math>8,16,24,32,\dots,2024\cong1,2,3,4,\dots,253</math>, which comprises <math>253</math> numbers. In the latter case, we have the numbers <math>7,11,15,19,\dots,2023\cong4,8,12,16,\dots,2020\cong1,2,3,4,\dots,505</math>, which comprises <math>505</math> numbers. There are <math>758</math> numbers in total, so our answer is <math>7+5+8=\boxed{\textbf{(D) 20}}</math>. |
~Technodoggo | ~Technodoggo |
Revision as of 11:48, 9 November 2024
- The following problem is from both the 2024 AMC 10A #18 and 2024 AMC 12A #11, so both problems redirect to this page.
Problem
There are exactly positive integers with such that the base- integer is divisible by (where is in base ten). What is the sum of the digits of ?
Solution 1
, if even then . If odd then so . Now so but is too small so . ~OronSH ~mathkiddus ~andliu766
Solution 2
Clearly, is either even or odd. If is even, let .
Thus, one solution is for some integer , or .
What if is odd? Then let :
This simply states that is odd. Thus, the other solution is for some integer , or .
We now simply must count the number of integers between and , inclusive, that are mod or mod . Note that the former case comprises even numbers only while the latter is only odd; thus, there is no overlap and we can safely count the number of each and add them.
In the former case, we have the numbers ; this list is equivalent to , which comprises numbers. In the latter case, we have the numbers , which comprises numbers. There are numbers in total, so our answer is .
~Technodoggo
Solution 3
Note that is to be divisible by , which means that is divisible by .
If , then is not divisible by .
If , then is not divisible by .
If , then is not divisible by .
If , then is divisible by .
If , then is not divisible by .
If , then is not divisible by .
If , then is divisible by .
If , then is divisible by .
Therefore, for every values of , of them will make divisible by . Therefore, since is divisible by , values of , but this includes , which does not satisfy the given inequality. Therefore, the answer is ~Tacos_are_yummy_1
Solution 4
We need take away one because is out of range, so
See also
2024 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.