Difference between revisions of "2018 AIME I Problems/Problem 2"
m (→Solution 2) |
(→Solution 1) |
||
Line 2: | Line 2: | ||
The number <math>n</math> can be written in base <math>14</math> as <math>\underline{a}\text{ }\underline{b}\text{ }\underline{c}</math>, can be written in base <math>15</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{b}</math>, and can be written in base <math>6</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }</math>, where <math>a > 0</math>. Find the base-<math>10</math> representation of <math>n</math>. | The number <math>n</math> can be written in base <math>14</math> as <math>\underline{a}\text{ }\underline{b}\text{ }\underline{c}</math>, can be written in base <math>15</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{b}</math>, and can be written in base <math>6</math> as <math>\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }</math>, where <math>a > 0</math>. Find the base-<math>10</math> representation of <math>n</math>. | ||
− | ==Solution== | + | ==Solution 1== |
We have these equations: | We have these equations: | ||
Line 9: | Line 9: | ||
Then we know <math>3a+b=22</math>. | Then we know <math>3a+b=22</math>. | ||
− | Taking the first two equations we see that <math>29a+ | + | Taking the first two equations we see that <math>29a+14c=13b</math>. Combining the two gives <math>a=4, c=10</math>. Then we see that <math>222 \times 4+37 \times1=\boxed{925}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 19:21, 31 December 2020
Contents
[hide]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
.
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.