Difference between revisions of "2025 AIME I Problems/Problem 1"
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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 | ||
− | ==Solution | + | ==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> | ||
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
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 2
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.