2025 AIME I Problems/Problem 1

Problem

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

Solution 1 (thorough)

We are tasked with finding the number of integer bases $b>9$ such that $\cfrac{9b+7}{b+7}\in\textbf{Z}$ . Notice that $\cfrac{9b+7}{b+7}=\cfrac{9b+63-56}{b+7}=\cfrac{9(b+7)-56}{b+7}=9-\cfrac{56}{b+7}$ so we need only $\cfrac{56}{b+7}\in\textbf{Z}$ . Then $b+7$ is a factor of $56$ .

The factors of $56$ are $1,2,4,7,8,14,28,56$ . Of these, only $8,14,28,56$ produce a positive $b$ , namely $b=1,7,21,49$ respectively. However, we are given that $b>9$ , so only $b=21,49$ are solutions. Thus the answer is $21+49=\boxed{070}$ . ~eevee9406

Solution 1 (quick)

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 $49+21=\boxed{070}.$

~mathkiddus

Solution 2

This means that $a(b+7)=9b+7$ where $a$ is a natural number. Rearranging we get $(a-9)(b+7)=-56$ . Since $b>9$ , $b=49,21$ . Thus the answer is $49+21=\boxed{70}$

~zhenghua

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2025 AIME I Problems/Problem 1

Contents

Problem

Solution 1 (thorough)

Solution 1 (quick)

Solution 2

See also