Difference between revisions of "2001 AIME I Problems/Problem 8"
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Since the <math>a_i</math>s are base-<math>7</math> digits, it follows that <math>a_i < 7</math>, and the LHS is less than or equal to <math>30</math>. Hence our number can have at most <math>3</math> digits in base-<math>7</math>. Letting <math>a_2 = 6</math>, we find that <math>630_7 = \boxed{315}_{10}</math> is our largest 7-10 double. | Since the <math>a_i</math>s are base-<math>7</math> digits, it follows that <math>a_i < 7</math>, and the LHS is less than or equal to <math>30</math>. Hence our number can have at most <math>3</math> digits in base-<math>7</math>. Letting <math>a_2 = 6</math>, we find that <math>630_7 = \boxed{315}_{10}</math> is our largest 7-10 double. | ||
+ | |||
+ | ==Solution 2 (Bash/Guess and Check)== | ||
+ | Let <math>A</math> be the base <math>10</math> representation of our number, and let <math>B</math> be its base <math>7</math> representation. | ||
+ | |||
+ | Given this is an AIME problem, <math>A<1000</math>. If we look at <math>B</math> in base <math>10</math>, it must be equal to <math>2A</math>, so <math>B<2000</math> when <math>B</math> is looked at in base <math>10.</math> | ||
+ | |||
+ | If <math>B</math> in base <math>10</math> is less than <math>2000</math>, then <math>B</math> as a number in base <math>7</math> must be less than <math>2*7^2=686</math>. | ||
+ | |||
+ | <math>686</math> is non-existent in base <math>7</math>, so we're gonna have to bump that down to <math>666_7</math>. | ||
+ | |||
+ | This suggests that <math>A</math> is less than <math>\frac{666}{2}=333</math>. | ||
+ | |||
+ | Guess and check shows that <math>310<A<320</math>, and checking values in that range produces <math>\boxed{315}_{10}</math>. | ||
== See also == | == See also == |
Revision as of 13:40, 9 May 2020
Problem
Call a positive integer a 7-10 double if the digits of the base-
representation of
form a base-
number that is twice
. For example,
is a 7-10 double because its base-
representation is
. What is the largest 7-10 double?
Solution
We let ; we are given that
(This is because the digits in
' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)
Expanding, we find that
or re-arranging,
Since the s are base-
digits, it follows that
, and the LHS is less than or equal to
. Hence our number can have at most
digits in base-
. Letting
, we find that
is our largest 7-10 double.
Solution 2 (Bash/Guess and Check)
Let be the base
representation of our number, and let
be its base
representation.
Given this is an AIME problem, . If we look at
in base
, it must be equal to
, so
when
is looked at in base
If in base
is less than
, then
as a number in base
must be less than
.
is non-existent in base
, so we're gonna have to bump that down to
.
This suggests that is less than
.
Guess and check shows that , and checking values in that range produces
.
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.