Difference between revisions of "2018 AIME I Problems/Problem 2"
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The problem is equivalent to finding a solution to the system of Diophantine equations <math>196a+14b+c=225a+15c+b</math> and <math>225a+15c+b=216a+36c+6a+c,</math> where <math>1\le a\le 5,\,0\le b\le 13,</math> and <math>0\le c\le 5.</math> Simplifying the second equation gives <math>b=22c-3a.</math> Substituting for <math>b</math> in the first equation and simplifying then gives <math>a=4c,</math> so <math>a = 4</math> and <math>c = 1,</math> and the base-<math>10</math> representation of <math>n</math> is <math>222 \cdot 4 + 37 \cdot 1 = 925.</math> It may be verified that <math>b=10\le 13.</math> | The problem is equivalent to finding a solution to the system of Diophantine equations <math>196a+14b+c=225a+15c+b</math> and <math>225a+15c+b=216a+36c+6a+c,</math> where <math>1\le a\le 5,\,0\le b\le 13,</math> and <math>0\le c\le 5.</math> Simplifying the second equation gives <math>b=22c-3a.</math> Substituting for <math>b</math> in the first equation and simplifying then gives <math>a=4c,</math> so <math>a = 4</math> and <math>c = 1,</math> and the base-<math>10</math> representation of <math>n</math> is <math>222 \cdot 4 + 37 \cdot 1 = 925.</math> It may be verified that <math>b=10\le 13.</math> | ||
==Solution 4 (Simple Modular Arithmetic)== | ==Solution 4 (Simple Modular Arithmetic)== | ||
− | We're given that <math>196a+14b+c=225a+15c+b= | + | We're given that <math>196a+14b+c=225a+15c+b=222a+37c.</math> By taking the difference of the first <math>2</math> equalities, we receive <math>29a+14c=13b.</math> Taking <math>\pmod{13}</math>, we receive <math>3a+c \equiv 0 \pmod{13}.</math> We receive the following cases: <math>(a,c)=(4,1)</math> or <math>(3,4).</math> (Note that <math>(2,7)</math> doesn't work since <math>a,c<6</math> by third condition). We can just check these two, and find that <math>(a,b,c)=(4,10,1),</math> and just plugging in <math>(a,c)</math> into the third expression we receive <math>888+37=\boxed{925}.</math> |
~SirAppel | ~SirAppel |
Latest revision as of 18:36, 21 February 2024
Contents
Problem
The number can be written in base
as
, can be written in base
as
, and can be written in base
as
, where
. Find the base-
representation of
.
Solution 1
We have these equations:
.
Taking the last two we get
. Because
otherwise
, and
,
.
Then we know .
Taking the first two equations we see that
. Combining the two gives
. Then we see that
.
Solution 2
We know that . Combining the first and third equations give that
, or
The second and third gives
, or
We can have
, but only
falls within the possible digits of base
. Thus
,
, and thus you can find
which equals
. Thus, our answer is
.
Solution 3 (Official MAA)
The problem is equivalent to finding a solution to the system of Diophantine equations and
where
and
Simplifying the second equation gives
Substituting for
in the first equation and simplifying then gives
so
and
and the base-
representation of
is
It may be verified that
Solution 4 (Simple Modular Arithmetic)
We're given that By taking the difference of the first
equalities, we receive
Taking
, we receive
We receive the following cases:
or
(Note that
doesn't work since
by third condition). We can just check these two, and find that
and just plugging in
into the third expression we receive
~SirAppel
Video Solution
https://www.youtube.com/watch?v=WVtbD8x9fCM ~Shreyas S
See Also
2018 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.