Difference between revisions of "2022 AIME I Problems/Problem 2"
MRENTHUSIASM (talk | contribs) m (→Solution 3) |
MRENTHUSIASM (talk | contribs) m (→Solution 2) |
||
Line 21: | Line 21: | ||
As shown in Solution 1, we get <math>99a = 71b+8c</math>. | As shown in Solution 1, we get <math>99a = 71b+8c</math>. | ||
− | Note that <math>99</math> and <math>71</math> are large numbers comparatively to <math>8</math>, 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 <math>99</math> and <math>71</math> is <math>28</math>, which is a multiple of <math>4</math>. So, if we multiply this by <math>2</math>, it will be a multiple of <math>8</math> and thus the gap can be filled. Therefore, | + | Note that <math>99</math> and <math>71</math> are large numbers comparatively to <math>8</math>, 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 <math>99</math> and <math>71</math> is <math>28</math>, which is a multiple of <math>4</math>. So, if we multiply this by <math>2</math>, it will be a multiple of <math>8</math> and thus the gap can be filled. Therefore, the only solution is <math>(a,b,c)=(2,2,7)</math>, and the answer is <math>\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}</math>. |
~KingRavi | ~KingRavi |
Revision as of 21:05, 19 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 large 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, the only solution is
, and the answer is
.
~KingRavi
Solution 3
As shown in Solution 1, we get
We list a few multiples of out:
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, we get .
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.