Difference between revisions of "2024 AMC 10A Problems/Problem 18"
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<math>2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8</math>, if <math>b</math> even then <math>b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8</math>. If <math>b</math> odd then <math>b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8</math> so <math>2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8</math>. Now <math>8\mid 2024</math> so <math>\tfrac38\cdot 2024=759</math> but <math>3</math> is too small so <math>758\implies\boxed{20}</math>. | <math>2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8</math>, if <math>b</math> even then <math>b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8</math>. If <math>b</math> odd then <math>b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8</math> so <math>2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8</math>. Now <math>8\mid 2024</math> so <math>\tfrac38\cdot 2024=759</math> but <math>3</math> is too small so <math>758\implies\boxed{20}</math>. | ||
~OronSH ~mathkiddus | ~OronSH ~mathkiddus | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | ==Solution 1== | ||
| + | |||
| + | \begin{align*} | ||
| + | 2024_b\equiv0\pmod{16} \\ | ||
| + | 2b^3+2b+4\equiv0\pmod{16} \\ | ||
| + | b^3+b+2\equiv0\pmod8 \\ | ||
| + | \end{align*} | ||
| + | |||
| + | Clearly, <math>b</math> is either even or odd. If <math>b</math> is even, let <math>b=2a</math>. | ||
| + | |||
| + | \begin{align*} | ||
| + | (2a)^3+2a+2\equiv0\pmod8 \\ | ||
| + | 8a^3+2a+2\equiv0\pmod8 \\ | ||
| + | 0+2a+2\equiv0\pmod8 \\ | ||
| + | a+1\equiv0\pmod4 \\ | ||
| + | a\equiv3\pmod4 \\ | ||
| + | \end{align*} | ||
| + | |||
| + | Thus, one solution is <math>b=2(4x+3)=8x+6</math> for some integer <math>x</math>, or <math>b\equiv6\pmod8</math>. | ||
| + | |||
| + | What if <math>b</math> is odd? Then let <math>b=2a+1</math>: | ||
| + | |||
| + | \begin{align*} | ||
| + | (2a+1)^3+2a+1+2\equiv0\pmod8 \\ | ||
| + | 8a^3+12a^2+6a+1+2a+1+2\equiv0\pmod8 \\ | ||
| + | 8a^3+12a^2+8a+4\equiv0\pmod8 \\ | ||
| + | 4a^2+4\equiv0\pmod8 \\ | ||
| + | a^2\equiv1\pmod2 \\ | ||
| + | \end{align*} | ||
| + | |||
| + | This simply states that <math>a</math> is odd. Thus, the other solution is <math>b=2(2x+1)+1=4x+3</math> for some integer <math>x</math>, or <math>b\equiv3\pmod4</math>. | ||
| + | |||
| + | 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\cong1,23,4,\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 | ||
| + | |||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2024|ab=A|num-b=17|num-a=19}} | ||
{{AMC12 box|year=2024|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2024|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:40, 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.
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
, if even then 
. If odd then 
so 
. Now 
so 
but 
is too small so 
.
~OronSH ~mathkiddus
Solution 2
Solution 1
\begin{align*} 2024_b\equiv0\pmod{16} \\ 2b^3+2b+4\equiv0\pmod{16} \\ b^3+b+2\equiv0\pmod8 \\ \end{align*}
Clearly, is either even or odd. If is even, let 
.
\begin{align*} (2a)^3+2a+2\equiv0\pmod8 \\ 8a^3+2a+2\equiv0\pmod8 \\ 0+2a+2\equiv0\pmod8 \\ a+1\equiv0\pmod4 \\ a\equiv3\pmod4 \\ \end{align*}
Thus, one solution is 
for some integer 
, or 
.
What if is odd? Then let 
:
\begin{align*} (2a+1)^3+2a+1+2\equiv0\pmod8 \\ 8a^3+12a^2+6a+1+2a+1+2\equiv0\pmod8 \\ 8a^3+12a^2+8a+4\equiv0\pmod8 \\ 4a^2+4\equiv0\pmod8 \\ a^2\equiv1\pmod2 \\ \end{align*}
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
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. 
