Difference between revisions of "2024 AMC 10A Problems/Problem 18"
(→Solution 3) |
(→Solution 3) |
||
Line 67: | Line 67: | ||
\end{align*} | \end{align*} | ||
− | by the way I am sorry I deleted the original solution | + | we need |
− | I wanted to add a new one actually | + | <math>b\ \equiv3\ (mod\ 8)\ \ or\ b\ \equiv6\ (mod\ 8)\ \ or\ b\ \equiv7\ (mod\ 8)</math> |
+ | <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 | ||
+ | 7+8+5=20 | ||
+ | |||
+ | 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. If anyone could recover the original solution feel free to get rid of mine. | ||
+ | |||
+ | ~xianghaoalexwang | ||
==See also== | ==See also== |
Revision as of 22:18, 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
we need take away one because 3 is out of range so 758 7+8+5=20
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. If anyone could recover the original solution feel free to get rid of mine.
~xianghaoalexwang
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.