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

(Solution 4)
(Solution 3)
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~Technodoggo
 
~Technodoggo
  
==Solution 4==
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==Solution 3==
  
<math>2024_b=2\ast\ b^3+2\ast\ b+4\ </math>
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<math>2024_b=2\ast\ b^3+2\ast\ b+4\ \
<math>2024_(b+8)=2\ast(b+8)^3+2\ast(b+8)+4</math>
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{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>
 
<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>
  
<math>2024_(b+8)-2024_b\equiv0\ (mod\ 16)</math>
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\begin{align*}
<math>2024_(b+8)\ \ \equiv2024_b\ \ (mod\ 16)</math>
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2024_(b+8)-2024_b\equiv0\ (mod\ 16)\
<math>2024_0\equiv4\ (mod\ 16)</math>
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2024_(b+8)\ \ \equiv2024_b\ \ (mod\ 16)\
<math>2024_1\equiv8\ (mod\ 16)</math>
+
2024_0\equiv4\ (mod\ 16)\
<math>2024_2\equiv6\ (mod\ 16)</math>
+
2024_1\equiv8\ (mod\ 16)\
<math>2024_3\equiv0(mod\ 16)</math>
+
2024_2\equiv6\ (mod\ 16)\
<math>2024_4\equiv12(mod\ 16)</math>
+
2024_3\equiv0(mod\ 16)\
<math>2024_5\equiv8(mod\ 16)</math>
+
2024_4\equiv12(mod\ 16)\
<math>2024_6\equiv0(mod\ 16)</math>
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2024_5\equiv8(mod\ 16)\
<math>2024_7\equiv0(mod\ 16)</math>
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2024_6\equiv0(mod\ 16)\
 +
2024_7\equiv0(mod\ 16)\
 +
\end{align*}
 +
 
 +
by the way I am sorry I deleted the original solution, how do I recover the original solution
 +
I wanted to add a new one actually
  
 
==See also==
 
==See also==

Revision as of 22:15, 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 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

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

Solution 3

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

2024(b+8)2024b0 (mod 16)2024(b+8)  2024b  (mod 16)202404 (mod 16)202418 (mod 16)202426 (mod 16)202430(mod 16)2024412(mod 16)202458(mod 16)202460(mod 16)202470(mod 16)

by the way I am sorry I deleted the original solution, how do I recover the original solution I wanted to add a new one actually

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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