Difference between revisions of "2022 AIME I Problems/Problem 2"
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== Solution 4 == | == Solution 4 == | ||
+ | As shown in Solution 1, the equation simplifies to <math>99a = 71b+8c</math>. | ||
+ | |||
+ | We can see that <math>99</math> is <math>28</math> larger than <math>71</math>, and we have an <math>8c</math>. We can clearly see that <math>56</math> is a multiple of <math>8</math>, and any larger than <math>56</math> would result in <math>c</math> being larger than <math>9</math>. Therefore, our only solution is <math>a = 2, b = 2, c = 7</math>. Our answer is <math>\boxed {227}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== |
Revision as of 23:31, 18 February 2022
Contents
Problem
Find the three-digit positive integer whose representation in base nine is
where
and
are (not necessarily distinct) digits.
Solution 1
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
As shown in Solution 1, we get . Note that
and
are big numbers comparatively to
, so we hypothesize that
and
are equal and
fills the gap between them. The difference between
and
is
, which is a multiple of
. So, if we multiply this by
, it will be a multiple of
and thus the gap can be filled. Therefore, a viable solution is
, and the answer is
.
~KingRavi
Solution 3
As shown in Solution 1, we get We list a few multiples of
out:
\begin{align*}
99\\198\\297\\396\\
\end{align*}
Of course,
can't be made of just
's. If we use one
, we get a remainder of
, which can't be made of
's either. So
doesn't work.
can't be made up of just
's. If we use one
, we get a remainder of
, which can't be made of
's. If we use two
's, we get a remainder of
, which can be made of
's.
Therefore we get
so
and
. Plugging this back into the original problem shows that this answer is indeed correct. Therefore,
~Technodoggo
Solution 4
As shown in Solution 1, the equation simplifies to .
We can see that is
larger than
, and we have an
. We can clearly see that
is a multiple of
, and any larger than
would result in
being larger than
. Therefore, our only solution is
. Our answer is
.
~Arcticturn
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=z5Y4bT5rL-s
Video Solution
https://www.youtube.com/watch?v=CwSkAHR3AcM
~Steven Chen (www.professorchenedu.com)
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.