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

Revision as of 14:06, 15 May 2025

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 2 (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 3

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

Solution 4

Let $\dfrac{9b+7}{b+7} = n$ . Now, we have: $\dfrac{9(b+7)-56}{b+7} = n \Longrightarrow 9-\dfrac{56}{b+7}$ . Now, we can just find the factors of $56$ , subtract $7$ , and sum them. Listing them out, we have the only ones that are positive are $8-1 = 7, 14-7 = 7, 28-7 = 21, 56-7 = 49$ . But, we have this condition: $b > 9$ , so the only ones that work are $21,49 \Longrightarrow 21 + 49 = \boxed{070}$

-jb2015007

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=J-0BapU4Yuk

Video Solution(Fast!, Easy, Beginner-Friendly)

https://www.youtube.com/watch?v=S8aakoJToM0

~MC

Video Solution by Mathletes Corner

https://www.youtube.com/watch?v=fEYpnDxSlk0

~GP102

Quick & Easy Video Solution

https://www.youtube.com/watch?v=A-h121roYg8

@@ Line 20: / Line 20: @@
 ~[[User:zhenghua|zhenghua]]
+==Solution 4==
+Let <math>\dfrac{9b+7}{b+7} = n</math>. Now, we have: <math>\dfrac{9(b+7)-56}{b+7} = n \Longrightarrow 9-\dfrac{56}{b+7}</math>. Now, we can just find the factors of <math>56</math>, subtract <math>7</math>, and sum them. Listing them out, we have the only ones that are positive are <math>8-1 = 7, 14-7 = 7, 28-7 = 21, 56-7 = 49</math>. But, we have this condition: <math>b > 9</math>, so the only ones that work are <math>21,49 \Longrightarrow 21 + 49 = \boxed{070}</math>
+-jb2015007
 ==Video Solution 1 by SpreadTheMathLove==

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