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

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==Solution 1==
 
==Solution 1==
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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>\frac38\cdot 2024=759</math> but one of the answers we got from that, <math>3</math>, is too small, so <math>759 - 1 = 758\implies\boxed{\textbf{(D) }20}</math>.
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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> is even then <math>b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8</math>. If <math>b</math> is 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 one of the answers we got from that, <math>3</math>, is too small, so <math>759 - 1 = 758\implies\boxed{\textbf{(D) }20}</math>.
  
 
~OronSH ~[[User:Mathkiddus|mathkiddus]] ~andliu766 ~megaboy6679 ~trevian1(minor edit for clarity)
 
~OronSH ~[[User:Mathkiddus|mathkiddus]] ~andliu766 ~megaboy6679 ~trevian1(minor edit for clarity)
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take away one because <math>b=3</math> is out of range, so <math>758\Rightarrow7+8+5=\boxed{\text{(D) }20}</math>
 
take away one because <math>b=3</math> is out of range, so <math>758\Rightarrow7+8+5=\boxed{\text{(D) }20}</math>
  
 +
===Solution 4.1: Alternate Ending===
  
−
==Video Solution by sharmaguy==  
+
One can arrange 2024 into different "blocks" consisting of 8 numbers each, particularly \( \{0,1,2,3,4,5,6,7\} \), \( \{8,9,10,11,12,13,14,15\} \) ...
 +
 
 +
These just represent different \( \mod(8) \) class residues. So, you can just divide 2024 by 8 blocks giving us 253 total blocks with 0 extra semi-blocks.
 +
 
 +
We see that \( 2024_3 \mod 8 \), \( 2024_6 \mod 8 \), and \( 2024_7 \mod 8 \) all appear exactly 253 times each, so we have \( 253 \times 3 = 759 \) different values of \( K \).
 +
 
 +
However, we must not include \( 2024_3 \), as the bases \( b \) range from \( 5 \le b \le 2024 \), therefore we have \( 759-1=758 \) cases.
 +
 
 +
Our answer is \( 7+5+8= \) <math>\boxed{\text{(D) }20}</math>.
 +
 
 +
~Pinotation (I did not do the solution above this remark)
 +
 
 +
==Solution 5 (Factoring)==
 +
 
 +
Start similar to other solutions.
 +
 
 +
<cmath>b^3+b+2 \equiv 0\pmod 8</cmath>
 +
 
 +
Except now we notice the following:
 +
 
 +
\begin{align*}
 +
b^3+8+b+2 \equiv 8\pmod 8 \equiv 0\pmod 8\\
 +
(b+2)(b^2-2b+4) + b+2 \equiv 0\pmod 8\\
 +
(b+2)(b^2-b+5) \equiv 0\pmod 8\\
 +
\end{align*}
 +
 
 +
Notice that <math>b+2</math> and <math>b^2-b+5</math> have opposite parity, and when multiplied, is divisible by <math>8</math>.
 +
Thus, either <math>b+2</math> or <math>b^2-b+5</math> is divisible by <math>8</math>.
 +
 
 +
<math>b+2</math> is obvious, where <math>b+2 | 8</math> when <math>b \equiv 6\pmod 8</math>
 +
 
 +
When testing for <math>b^2-b+5 | 8</math>, we find <math>b \equiv 3, 7\pmod 8</math>
 +
 
 +
Proceed similar to the other solutions, <math>\lfloor\dfrac{3}{8} \cdot 2024\rfloor-1=759-1=758</math>, where <math>7+5+8=\boxed{\text{(D) }20}</math>
 +
 
 +
~ NSAoPS
 +
 
 +
==Video Solution by [https://artofproblemsolving.com/wiki/index.php/User:grogg007 grogg007]==  
 
https://youtu.be/xDye0tT4oPc
 
https://youtu.be/xDye0tT4oPc
  
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https://youtu.be/snp3Yo8rU-M?si=GHqpGqQs6Q0xM57j
 
https://youtu.be/snp3Yo8rU-M?si=GHqpGqQs6Q0xM57j
  
−
==See also==
+
== Video solution by TheNeuralMathAcademy ==
−
{{AMC10 box|year=2024|ab=A|num-b=17|num-a=19}}
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https://www.youtube.com/watch?v=4b_YLnyegtw&t=3726s
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{{AMC12 box|year=2024|ab=A|num-b=10|num-a=12}}
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 +
==See Also==
 +
{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
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* [[AMC 10]]
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* [[AMC 10 Problems and Solutions]]
 +
* [[Mathematics competitions]]
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* [[Mathematics competition resources]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:47, 28 September 2025

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 $b$ is even then $b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8$. If $b$ is 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 one of the answers we got from that, $3$, is too small, so $759 - 1 = 758\implies\boxed{\textbf{(D) }20}$.

~OronSH ~mathkiddus ~andliu766 ~megaboy6679 ~trevian1(minor edit for clarity)

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 $758\Rightarrow7+8+5=\boxed{\text{(D) }20}$

Solution 4.1: Alternate Ending

One can arrange 2024 into different "blocks" consisting of 8 numbers each, particularly \( \{0,1,2,3,4,5,6,7\} \), \( \{8,9,10,11,12,13,14,15\} \) ...

These just represent different \( \mod(8) \) class residues. So, you can just divide 2024 by 8 blocks giving us 253 total blocks with 0 extra semi-blocks.

We see that \( 2024_3 \mod 8 \), \( 2024_6 \mod 8 \), and \( 2024_7 \mod 8 \) all appear exactly 253 times each, so we have \( 253 \times 3 = 759 \) different values of \( K \).

However, we must not include \( 2024_3 \), as the bases \( b \) range from \( 5 \le b \le 2024 \), therefore we have \( 759-1=758 \) cases.

Our answer is \( 7+5+8= \) $\boxed{\text{(D) }20}$.

~Pinotation (I did not do the solution above this remark)

Solution 5 (Factoring)

Start similar to other solutions.

\[b^3+b+2 \equiv 0\pmod 8\]

Except now we notice the following:

\begin{align*} b^3+8+b+2 \equiv 8\pmod 8 \equiv 0\pmod 8\\ (b+2)(b^2-2b+4) + b+2 \equiv 0\pmod 8\\ (b+2)(b^2-b+5) \equiv 0\pmod 8\\ \end{align*}

Notice that $b+2$ and $b^2-b+5$ have opposite parity, and when multiplied, is divisible by $8$. Thus, either $b+2$ or $b^2-b+5$ is divisible by $8$.

$b+2$ is obvious, where $b+2 | 8$ when $b \equiv 6\pmod 8$

When testing for $b^2-b+5 | 8$, we find $b \equiv 3, 7\pmod 8$

Proceed similar to the other solutions, $\lfloor\dfrac{3}{8} \cdot 2024\rfloor-1=759-1=758$, where $7+5+8=\boxed{\text{(D) }20}$

~ NSAoPS

Video Solution by grogg007

https://youtu.be/xDye0tT4oPc

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://youtu.be/snp3Yo8rU-M?si=GHqpGqQs6Q0xM57j

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=3726s

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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All AMC 10 Problems and Solutions

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