Difference between revisions of "2025 AIME I Problems/Problem 1"
Mathkiddus (talk | contribs) (→Solution 1) |
|||
| Line 5: | Line 5: | ||
~[[User:Mathkiddus|mathkiddus]] | ~[[User:Mathkiddus|mathkiddus]] | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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]] | ||
Revision as of 19:20, 13 February 2025
Problem
Find the sum of all integer bases 
for which 
is a divisor of 
Solution 1
We have, 
meaning 
so taking divisors of 
under bounds to find 
meaning our answer is 
Solution 2
This means that 
where 
is a natural number. Rearranging we get 
. Since , 
. Thus the answer is 
