Difference between revisions of "2025 AIME I Problems/Problem 1"

(Problem)
(Solution 1)
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Find the sum of all integer bases <math>b>9</math> for which <math>17_b</math> is a divisor of <math>97_b.</math>
 
Find the sum of all integer bases <math>b>9</math> for which <math>17_b</math> is a divisor of <math>97_b.</math>
 
==Solution 1==
 
==Solution 1==
We have, <math>b + 7 \mid 9b + 7</math> meaning <math>b + 7 \mid -56</math> so taking divisors of 56 under bounds to find <math>b = 49, 21</math> meaning our answer is <math>49+21=\boxed{070}.</math>
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We have, <math>b + 7 \mid 9b + 7</math> meaning <math>b + 7 \mid -56</math> so taking divisors of <math>56</math> under bounds to find <math>b = 49, 21</math> meaning our answer is <math>49+21=\boxed{070}.</math>

Revision as of 16:57, 13 February 2025

Problem

Find the sum of all integer bases $b>9$ for which $17_b$ is a divisor of $97_b.$

Solution 1

We have, $b + 7 \mid 9b + 7$ meaning $b + 7 \mid -56$ so taking divisors of $56$ under bounds to find $b = 49, 21$ meaning our answer is $49+21=\boxed{070}.$