Difference between revisions of "2024 AMC 10A Problems/Problem 18"
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~Technodoggo | ~Technodoggo | ||
− | ==Solution | + | ==Solution 3== |
− | <math>2024_b=2\ast\ b^3+2\ast\ b+4\ | + | <math>2024_b=2\ast\ b^3+2\ast\ b+4\ \ |
− | + | {2024}_{\left(b+8\right)}=2\ast\left(b+8\right)^3+2\ast\left(b+8\right)+4</math> | |
<math>{2024}_{\left(b+8\right)}-{2024}_b=2*\left(8\right)*\left(b^2+8b+64\right)+2*8\ =16*\left(b^2+8b+64\right)+16</math> | <math>{2024}_{\left(b+8\right)}-{2024}_b=2*\left(8\right)*\left(b^2+8b+64\right)+2*8\ =16*\left(b^2+8b+64\right)+16</math> | ||
− | + | \begin{align*} | |
− | + | 2024_(b+8)-2024_b\equiv0\ (mod\ 16)\ | |
− | + | 2024_(b+8)\ \ \equiv2024_b\ \ (mod\ 16)\ | |
− | + | 2024_0\equiv4\ (mod\ 16)\ | |
− | + | 2024_1\equiv8\ (mod\ 16)\ | |
− | + | 2024_2\equiv6\ (mod\ 16)\ | |
− | + | 2024_3\equiv0(mod\ 16)\ | |
− | + | 2024_4\equiv12(mod\ 16)\ | |
− | + | 2024_5\equiv8(mod\ 16)\ | |
− | + | 2024_6\equiv0(mod\ 16)\ | |
+ | 2024_7\equiv0(mod\ 16)\ | ||
+ | \end{align*} | ||
+ | |||
+ | by the way I am sorry I deleted the original solution, how do I recover the original solution | ||
+ | I wanted to add a new one actually | ||
==See also== | ==See also== |
Revision as of 22:15, 8 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.
Contents
[hide]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
by the way I am sorry I deleted the original solution, how do I recover the original solution I wanted to add a new one actually
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.