Difference between revisions of "2025 AIME I Problems/Problem 1"
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==Problem== | ==Problem== | ||
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== | + | |
| + | ==Video solution by [[User:grogg007|grogg007]]== | ||
| + | https://www.youtube.com/watch?v=PNBxBvvjbcU | ||
| + | |||
| + | ==Solution 1 (thorough)== | ||
| + | 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 | ||
| + | <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>. | ||
| + | |||
| + | |||
| + | 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 | ||
| + | |||
| + | ==Solution 2 (quick)== | ||
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> | 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> | ||
~[[User:Mathkiddus|mathkiddus]] | ~[[User:Mathkiddus|mathkiddus]] | ||
| − | ==Solution | + | ==Solution 3== |
| − | 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{ | + | 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{070}</math> |
~[[User:zhenghua|zhenghua]] | ~[[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== | ||
| + | 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== | ||
| + | {{AIME box|year=2025|before=First Problem|num-a=2|n=I}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 21:49, 16 July 2025
Contents
- 1 Problem
- 2 Video solution by grogg007
- 3 Solution 1 (thorough)
- 4 Solution 2 (quick)
- 5 Solution 3
- 6 Solution 4
- 7 Video Solution 1 by SpreadTheMathLove
- 8 Video Solution by Steakmath (simplest)
- 9 Video Solution(Fast!, Easy, Beginner-Friendly)
- 10 Video Solution by Mathletes Corner
- 11 Quick & Easy Video Solution
- 12 See also
Problem
Find the sum of all integer bases 
for which 
is a divisor of 
Video solution by grogg007
https://www.youtube.com/watch?v=PNBxBvvjbcU
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 
Solution 4
Let 
. Now, we have: 
. Now, we can just find the factors of , subtract 
, and sum them. Listing them out, we have the only ones that are positive are 
. But, we have this condition: 
, so the only ones that work are 
-jb2015007
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=J-0BapU4Yuk
Video Solution by Steakmath (simplest)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
