2024 AMC 10A Problems/Problem 18

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 $759 - 1 = 758\implies\boxed{20}$ . ~OronSH ~mathkiddus ~andliu766

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 $b^3+b+2$ is not divisible by $8$ .

If $b=1$ , then $b^3+b+2$ is not divisible by $8$ .

If $b=2$ , then $b^3+b+2$ is not divisible by $8$ .

If $b=3$ , then $b^3+b+2$ is divisible by $8$ .

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

If $b=5$ , then $b^3+b+2$ is not divisible by $8$ .

If $b=6$ , then $b^3+b+2$ is divisible by $8$ .

If $b=7$ , then $b^3+b+2$ 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

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 Pi Academy

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

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 10 Problems and Solutions

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 12 Problems and Solutions

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