Difference between revisions of "2022 AIME I Problems/Problem 2"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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+ | == Solution 2 (Simple) == | ||
+ | |||
+ | As in the previous solution, we get <math>99a = 71b+8c</math>. 99 and 71 are big numbers comparatively to 8, so we hypothesize that <math>a</math> and <math>b</math> are equal and <math>8c</math> fills the gap between them. The difference between 99 and 71 is 28, which is a multiple of 4, so if we multiply this by 2, it will be a multiple of 8 and thus the gap can be filled. Therefore, a viable solution is <math>a = 2, b = 2, c = 7</math> and the answer is <math>\boxed{227}</math> | ||
+ | |||
+ | ~KingRavi | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== |
Revision as of 20:01, 17 February 2022
Contents
Problem
Find the three-digit positive integer whose representation in base nine is
where
and
are (not necessarily distinct) digits.
Solution
We are given that which rearranges to
Taking both sides modulo
we have
The only solution occurs at
from which
Therefore, the requested three-digit positive integer is
~MRENTHUSIASM
Solution 2 (Simple)
As in the previous solution, we get . 99 and 71 are big numbers comparatively to 8, so we hypothesize that
and
are equal and
fills the gap between them. The difference between 99 and 71 is 28, which is a multiple of 4, so if we multiply this by 2, it will be a multiple of 8 and thus the gap can be filled. Therefore, a viable solution is
and the answer is
~KingRavi
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=z5Y4bT5rL-s
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.