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

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m (→‎Solution 1 (thorough))
 
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==Solution 1 (thorough)==
 
==Solution 1 (thorough)==
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We are tasked with finding the number of integer bases <math>b>9</math> such that <math>\cfrac{9b+7}{b+7}\in\textbf{Z}</math>. Notice that
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We are tasked with finding the number of integer bases <math>b>9</math> such that <math>\cfrac{9b+7}{b+7}\in\mathbb{Z}</math>. Notice that
 
<cmath>\cfrac{9b+7}{b+7}=\cfrac{9b+63-56}{b+7}=\cfrac{9(b+7)-56}{b+7}=9-\cfrac{56}{b+7}</cmath>
 
<cmath>\cfrac{9b+7}{b+7}=\cfrac{9b+63-56}{b+7}=\cfrac{9(b+7)-56}{b+7}=9-\cfrac{56}{b+7}</cmath>
−
so we need only <math>\cfrac{56}{b+7}\in\textbf{Z}</math>. Then <math>b+7</math> is a factor of <math>56</math>.
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so we need only <math>\cfrac{56}{b+7}\in\mathbb{Z}</math>. Then <math>b+7</math> is a factor of <math>56</math>.
  
  
−
The factors of <math>56</math> are <math>1,2,4,7,8,14,28,56</math>. Of these, only <math>8,14,28,56</math> produce a positive <math>b</math>, namely <math>b=1,7,21,49</math> respectively. However, we are given that <math>b>9</math>, so only <math>b=21,49</math> are solutions. Thus the answer is <math>21+49=\boxed{070}</math>. ~eevee9406
+
The factors of <math>56</math> are <math>1,2,4,7,8,14,28,56</math>. Of these, only <math>8,14,28,56</math> produce a positive <math>b</math>, namely <math>b=1,7,21,49</math> respectively. However, we are given that <math>b>9</math>, so only <math>b=21,49</math> are solutions. Thus the answer is <math>21+49=\boxed{070}</math>. ~eevee9406 ~NOOK(Minor LaTeX Edits)
  
 
==Solution 2 (quick)==
 
==Solution 2 (quick)==

Latest revision as of 15:26, 31 October 2025

Problem

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

Video solution by grogg007

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

Solution 1 (thorough)

We are tasked with finding the number of integer bases $b>9$ such that $\cfrac{9b+7}{b+7}\in\mathbb{Z}$. 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 $\cfrac{56}{b+7}\in\mathbb{Z}$. 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 ~NOOK(Minor LaTeX Edits)

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 $49+21=\boxed{070}$

~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

Solution 5 (Solution 4 but different approach)

We want \( 17_b \) to divide \( 97_b \). Converting to base 10 gives \( 17_b = b + 7 \) and \( 97_b = 9b + 7 \). The condition is \( b + 7 \mid 9b + 7 \). Subtracting \( 9(b + 7) \) from \( 9b + 7 \) gives \( (9b + 7) - 9(b + 7) = -56 \). So \( b + 7 \) must divide 56. Continue as in Solution 4 to get $\boxed{070}$

~Pinotation

Video Solution 1 by SpreadTheMathLove

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

Video Solution by Steakmath (simplest)

https://youtu.be/Qi8EjzfoLUU

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

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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