2024 AMC 10A Problems/Problem 18: Difference between revisions

Revision as of 17:27, 8 November 2024

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

2024 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 ~OronSH ~mathkiddus
 ==See also==
-{{AMC10 box|year=2024|ab=A|before=17|num-a=19}}
+{{AMC10 box|year=2024|ab=A|num-b=17|num-a=19}}
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2024 AMC 10A Problems/Problem 18: Difference between revisions

Revision as of 17:27, 8 November 2024

Problem

Solution

See also