Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10A Problems/Problem 18"

(→‎Problem)
(→‎Solution 3)
 
(24 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2024 AMC 10A Problems/Problem 18|2024 AMC 10A #18]] and [[2024 AMC 12A Problems/Problem 11|2024 AMC 12A #11]]}}
 +
 
==Problem==
 
==Problem==
 
There are exactly <math>K</math> positive integers <math>b</math> with <math>5 \leq b \leq 2024</math> such that the base-<math>b</math> integer <math>2024_b</math> is divisible by <math>16</math> (where <math>16</math> is in base ten). What is the sum of the digits of <math>K</math>?
 
There are exactly <math>K</math> positive integers <math>b</math> with <math>5 \leq b \leq 2024</math> such that the base-<math>b</math> integer <math>2024_b</math> is divisible by <math>16</math> (where <math>16</math> is in base ten). What is the sum of the digits of <math>K</math>?
Line 4: Line 6:
 
<math>\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad\textbf{(E) }21</math>
 
<math>\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad\textbf{(E) }21</math>
  
−
==Solution==
+
==Solution 1==
 +
<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>\frac38\cdot 2024=759</math> but <math>3</math> is too small so <math>759 - 1 = 758\implies\boxed{\textbf{(D) }20}</math>.
 +
 
 +
~OronSH ~[[User:Mathkiddus|mathkiddus]] ~andliu766 ~megaboy6679
 +
 
 +
==Solution 2==
 +
 
 +
\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\cong1,2,3,4,\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
 +
 
 +
==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 \equiv (0)^3 + (0) + 2 \equiv 2</math> is not divisible by <math>8</math>.
 +
 
 +
If <math>b=1</math>, then <math>b^3+b+2 \equiv (1)^3 + (1) + 2  \equiv 4</math> is not divisible by <math>8</math>.
 +
 
 +
If <math>b=2</math>, then <math>b^3+b+2 \equiv (2)^3 + (2) + 2  \equiv 4</math> is not divisible by <math>8</math>.
 +
 
 +
If <math>b=3</math>, then <math>b^3+b+2 \equiv (3)^3 + (3) + 2  \equiv (8+1)\cdot3 + (3) + 2 \equiv 8 </math> is divisible by <math>8</math>.
 +
 
 +
If <math>b=4</math>, then <math>b^3+b+2 \equiv (4)^3 + (4) + 2 \equiv 0 + 4 + 2 \equiv 6 </math> is not divisible by <math>8</math>.
 +
 
 +
If <math>b=5</math>, then <math>b^3+b+2 \equiv (-3)^3 + (-3) + 2 \equiv  (8+1)\cdot 3 + (-3) + 2 \equiv 2 </math> is not divisible by <math>8</math>.
 +
 
 +
If <math>b=6</math>, then <math>b^3+b+2 \equiv (-2)^3 + (-2) + 2 \equiv -8 + (-2) + 2 \equiv 0 </math> is divisible by <math>8</math>.
 +
 
 +
If <math>b=7</math>, then <math>b^3+b+2 \equiv (-1)^3 + (-1) + 2 \equiv -1 + (-1) + 2 \equiv 0 </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
 +
 
 +
More detail by ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 
 +
==Solution 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>
 +
 
 +
\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*}
 +
 
 +
We need
 +
<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>
 +
take away one because <math>b=3</math> is out of range, so <math>758\Rightarrow7+8+5=\boxed{\text{(D) }20}</math>
 +
 
 +
 
 +
== Video Solution by Power Solve ==
 +
https://www.youtube.com/watch?v=qtFvaD9TEaA
 +
 
 +
== Video Solution by Pi Academy ==
 +
 
 +
https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=6SQ74nt3ynw
 +
 
 +
==See also==
 +
{{AMC10 box|year=2024|ab=A|num-b=17|num-a=19}}
 +
{{AMC12 box|year=2024|ab=A|num-b=10|num-a=12}}
 +
{{MAA Notice}}

Latest revision as of 14:37, 17 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 $K$ positive integers $b$ with $5 \leq b \leq 2024$ such that the base-$b$ integer $2024_b$ is divisible by $16$ (where $16$ is in base ten). What is the sum of the digits of $K$?

$\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad\textbf{(E) }21$

Solution 1

$2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8$, if $b$ even then $b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8$. If $b$ odd then $b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8$ so $2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8$. Now $8\mid 2024$ so $\frac38\cdot 2024=759$ but $3$ is too small so $759 - 1 = 758\implies\boxed{\textbf{(D) }20}$.

~OronSH ~mathkiddus ~andliu766 ~megaboy6679

Solution 2

\begin{align*} 2024_b\equiv0\pmod{16} \\ 2b^3+2b+4\equiv0\pmod{16} \\ b^3+b+2\equiv0\pmod8 \\ \end{align*}

Clearly, $b$ is either even or odd. If $b$ is even, let $b=2a$.

\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 $b=2(4x+3)=8x+6$ for some integer $x$, or $b\equiv6\pmod8$.

What if $b$ is odd? Then let $b=2a+1$:

\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 $a$ is odd. Thus, the other solution is $b=2(2x+1)+1=4x+3$ for some integer $x$, or $b\equiv3\pmod4$.

We now simply must count the number of integers between $5$ and $2024$, inclusive, that are $6$ mod $8$ or $3$ mod $4$. 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 $6,14,22,30,\dots,2022$; this list is equivalent to $8,16,24,32,\dots,2024\cong1,2,3,4,\dots,253$, which comprises $253$ numbers. In the latter case, we have the numbers $7,11,15,19,\dots,2023\cong4,8,12,16,\dots,2020\cong1,2,3,4,\dots,505$, which comprises $505$ numbers. There are $758$ numbers in total, so our answer is $7+5+8=\boxed{\textbf{(D) 20}}$.

~Technodoggo

Solution 3

Note that $2024_b=2b^3+2b+4$ is to be divisible by $16$, which means that $b^3+b+2$ is divisible by $8$.

If $b=0$, then $b^3+b+2 \equiv (0)^3 + (0) + 2 \equiv 2$ is not divisible by $8$.

If $b=1$, then $b^3+b+2 \equiv (1)^3 + (1) + 2 \equiv 4$ is not divisible by $8$.

If $b=2$, then $b^3+b+2 \equiv (2)^3 + (2) + 2 \equiv 4$ is not divisible by $8$.

If $b=3$, then $b^3+b+2 \equiv (3)^3 + (3) + 2 \equiv (8+1)\cdot3 + (3) + 2 \equiv 8$ is divisible by $8$.

If $b=4$, then $b^3+b+2 \equiv (4)^3 + (4) + 2 \equiv 0 + 4 + 2 \equiv 6$ is not divisible by $8$.

If $b=5$, then $b^3+b+2 \equiv (-3)^3 + (-3) + 2 \equiv (8+1)\cdot 3 + (-3) + 2 \equiv 2$ is not divisible by $8$.

If $b=6$, then $b^3+b+2 \equiv (-2)^3 + (-2) + 2 \equiv -8 + (-2) + 2 \equiv 0$ is divisible by $8$.

If $b=7$, then $b^3+b+2 \equiv (-1)^3 + (-1) + 2 \equiv -1 + (-1) + 2 \equiv 0$ is divisible by $8$.

Therefore, for every $8$ values of $b$, $3$ of them will make $b^3+b+2$ divisible by $8$. Therefore, since $2024$ is divisible by $8$, $\dfrac{3}{8}\cdot2024=759$ values of $b$, but this includes $b=3$, which does not satisfy the given inequality. Therefore, the answer is \[759-1=758\rightarrow7+5+8=\boxed{\text{(D) }20}\] ~Tacos_are_yummy_1

More detail by ~luckuso

Solution 4

$2024_b=2\ast\ b^3+2\ast\ b+4\ \\ {2024}_{\left(b+8\right)}=2\ast\left(b+8\right)^3+2\ast\left(b+8\right)+4$ ${2024}_{\left(b+8\right)}-{2024}_b=2*\left(8\right)*\left(b^2+8b+64\right)+2*8\ =16*\left(b^2+8b+64\right)+16$

\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*}

We need $b\ \equiv3\ (mod\ 8)\ \ or\ b\ \equiv6\ (mod\ 8)\ \ or\ b\ \equiv7\ (mod\ 8) \\ \lfloor(2024-3)/8\rfloor+\lfloor(2024-6)/8\rfloor+\lfloor(2024-7)/8\rfloor+3=759$ take away one because $b=3$ is out of range, so $758\Rightarrow7+8+5=\boxed{\text{(D) }20}$


Video Solution by Power Solve

https://www.youtube.com/watch?v=qtFvaD9TEaA

Video Solution by Pi Academy

https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_18&oldid=234688"