Difference between revisions of "2025 AIME I Problems/Problem 1"
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This means that <math>a(b+7)=9b+7</math> where <math>a</math> is a natural number. Rearranging we get <math>(a-9)(b+7)=-56</math>. Since <math>b>9</math>, <math>b=49,21</math>. Thus the answer is <math>49+21=\boxed{70}</math> | This means that <math>a(b+7)=9b+7</math> where <math>a</math> is a natural number. Rearranging we get <math>(a-9)(b+7)=-56</math>. Since <math>b>9</math>, <math>b=49,21</math>. Thus the answer is <math>49+21=\boxed{70}</math> | ||
~[[User:zhenghua|zhenghua]] | ~[[User:zhenghua|zhenghua]] | ||
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==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=J-0BapU4Yuk | https://www.youtube.com/watch?v=J-0BapU4Yuk | ||
Latest revision as of 23:47, 23 February 2025
Contents
Problem
Find the sum of all integer bases 
for which 
is a divisor of 
Solution 1 (thorough)
We are tasked with finding the number of integer bases such that 
. Notice that
![\[\cfrac{9b+7}{b+7}=\cfrac{9b+63-56}{b+7}=\cfrac{9(b+7)-56}{b+7}=9-\cfrac{56}{b+7}\]](http://latex.artofproblemsolving.com/2/8/9/2892d2f999456eaba99d20297fb3d79052db249b.png)
so we need only 
. Then 
is a factor of 
.
The factors of are 
. Of these, only 
produce a positive 
, namely 
respectively. However, we are given that , so only 
are solutions. Thus the answer is 
. ~eevee9406
Solution 2 (quick)
We have, 
meaning 
so taking divisors of under bounds to find 
meaning our answer is 
Solution 3
This means that 
where 
is a natural number. Rearranging we get 
. Since , 
. Thus the answer is 
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
