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Difference between revisions of "2024 AMC 10A Problems/Problem 18"
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{{duplicate|[[2024 AMC 10A Problems/Problem 18|2024 AMC 10A #18]] and [[2024 AMC 12A Problems/Problem 11|2024 AMC 12A #11]]}}
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==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>?
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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}}
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{{AMC12 box|year=2024|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:36, 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

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

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