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
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 +
==Solution 2==
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 +
==Solution 1==
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 +
\begin{align*}
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2024_b\equiv0\pmod{16} \
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2b^3+2b+4\equiv0\pmod{16} \
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b^3+b+2\equiv0\pmod8 \
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\end{align*}
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Clearly, <math>b</math> is either even or odd. If <math>b</math> is even, let <math>b=2a</math>.
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\begin{align*}
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(2a)^3+2a+2\equiv0\pmod8 \
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8a^3+2a+2\equiv0\pmod8 \
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0+2a+2\equiv0\pmod8 \
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a+1\equiv0\pmod4 \
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a\equiv3\pmod4 \
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\end{align*}
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Thus, one solution is <math>b=2(4x+3)=8x+6</math> for some integer <math>x</math>, or <math>b\equiv6\pmod8</math>.
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What if <math>b</math> is odd? Then let <math>b=2a+1</math>:
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\begin{align*}
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(2a+1)^3+2a+1+2\equiv0\pmod8 \
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8a^3+12a^2+6a+1+2a+1+2\equiv0\pmod8 \
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8a^3+12a^2+8a+4\equiv0\pmod8 \
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4a^2+4\equiv0\pmod8 \
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a^2\equiv1\pmod2 \
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\end{align*}
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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>.
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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.
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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>.
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~Technodoggo
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==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 $K$ 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

$2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8$, 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 $\tfrac38\cdot 2024=759$ but $3$ is too small so $758\implies\boxed{20}$. ~OronSH ~mathkiddus

Solution 2

Solution 1

2024b0(mod16)2b3+2b+40(mod16)b3+b+20(mod8)

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

(2a)3+2a+20(mod8)8a3+2a+20(mod8)0+2a+20(mod8)a+10(mod4)a3(mod4)

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 $b=2a+1$:

(2a+1)3+2a+1+20(mod8)8a3+12a2+6a+1+2a+1+20(mod8)8a3+12a2+8a+40(mod8)4a2+40(mod8)a21(mod2)

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 $8,16,24,32,\dots,2024\cong1,23,4,\dots,253$, 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

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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