Difference between revisions of "2024 AMC 10A Problems/Problem 18"
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==Solution 3== | ==Solution 3== | ||
+ | Note that <math>2024_b=2b^3+2b+4</math> is to be divisible by <math>16</math>, which means that <math>b^3+b+2</math> is divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=0</math>, then <math>b^3+b+2</math> is not divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=1</math>, then <math>b^3+b+2</math> is not divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=2</math>, then <math>b^3+b+2</math> is not divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=3</math>, then <math>b^3+b+2</math> is divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=4</math>, then <math>b^3+b+2</math> is not divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=5</math>, then <math>b^3+b+2</math> is not divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=6</math>, then <math>b^3+b+2</math> is divisible by <math>8</math>. | ||
+ | |||
+ | If <math>b=7</math>, then <math>b^3+b+2</math> is divisible by <math>8</math>. | ||
+ | |||
+ | Therefore, for every <math>8</math> values of <math>b</math>, <math>3</math> of them will make <math>b^3+b+2</math> divisible by <math>8</math>. Therefore, since <math>2024</math> is divisible by <math>8</math>, <math>\dfrac{3}{8}\cdot2024=759</math> values of <math>b</math>, but this includes <math>b=3</math>, which does not satisfy the given inequality. Therefore, the answer is <cmath>759-1=758\rightarrow7+5+8=\boxed{\text{(D) }20}</cmath> ~Tacos_are_yummy_1 | ||
+ | |||
+ | ==Solution 4== | ||
<math>2024_b=2\ast\ b^3+2\ast\ b+4\ \ | <math>2024_b=2\ast\ b^3+2\ast\ b+4\ \ | ||
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\begin{align*} | \begin{align*} | ||
− | 2024_(b+8)-2024_b\equiv0\ (mod\ 16)\ | + | 2024_{(b+8)}-2024_b\equiv0\ (mod\ 16)\ |
− | 2024_(b+8)\ \ \equiv2024_b\ \ (mod\ 16)\ | + | 2024_{(b+8)}\ \ \equiv2024_b\ \ (mod\ 16)\ |
2024_0\equiv4\ (mod\ 16)\ | 2024_0\equiv4\ (mod\ 16)\ | ||
2024_1\equiv8\ (mod\ 16)\ | 2024_1\equiv8\ (mod\ 16)\ | ||
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\end{align*} | \end{align*} | ||
− | + | We need | |
<math>b\ \equiv3\ (mod\ 8)\ \ or\ b\ \equiv6\ (mod\ 8)\ \ or\ b\ \equiv7\ (mod\ 8) \ | <math>b\ \equiv3\ (mod\ 8)\ \ or\ b\ \equiv6\ (mod\ 8)\ \ or\ b\ \equiv7\ (mod\ 8) \ | ||
\lfloor(2024-3)/8\rfloor+\lfloor(2024-6)/8\rfloor+\lfloor(2024-7)/8\rfloor+3=759</math> | \lfloor(2024-3)/8\rfloor+\lfloor(2024-6)/8\rfloor+\lfloor(2024-7)/8\rfloor+3=759</math> | ||
− | take away one because 3 is out of range so 758 | + | take away one because <math>b=3</math> is out of range, so <math>758\Rightarrow7+8+5=\boxed{\text{(D) }20}</math> |
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==See also== | ==See also== |
Revision as of 11:46, 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.