Difference between revisions of "2001 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
+ | Call a positive integer <math>N</math> a ''7-10 double'' if the digits of the base-<math>7</math> representation of <math>N</math> form a base-<math>10</math> number that is twice <math>N</math>. For example, <math>51</math> is a 7-10 double because its base-<math>7</math> representation is <math>102</math>. What is the largest 7-10 double? | ||
== Solution == | == Solution == | ||
+ | We let <math>N_7 = \overline{a_na_{n-1}\cdots a_0}_7</math>; we are given that | ||
+ | |||
+ | <cmath>2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}</cmath> (This is because the digits in <math>N</math> ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2) | ||
+ | |||
+ | Expanding, we find that | ||
+ | |||
+ | <cmath>2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0</cmath> | ||
+ | |||
+ | or re-arranging, | ||
+ | |||
+ | <cmath>a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n</cmath> | ||
+ | |||
+ | 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 (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*7^3=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>A<320</math>, and checking values in that range produces <math>\boxed{315}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be <cmath>abc</cmath> in base 7. Then the number in expanded form is <cmath>49a+7b+c</cmath> in base 7 and <cmath>100a+10b+c</cmath> in base 10. Since the number in base 7 is half the number in base 10, we get the following equation. | ||
+ | <cmath>98a+14b+2c=100a+10b+c</cmath> which simplifies to <cmath>2a=4b+c.</cmath> | ||
+ | The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, <math>b</math> is <math>3</math> and <math>c</math> is <math>0</math>. Therefore, the largest 7-10 double is 630 in base 7, or <math>\boxed{315}</math> in base 10. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2001|n=I|num-b=7|num-a=9}} | |
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 11:28, 9 December 2023
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 (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
.
Solution 3
Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be in base 7. Then the number in expanded form is
in base 7 and
in base 10. Since the number in base 7 is half the number in base 10, we get the following equation.
which simplifies to
The largest possible value of a is 6 because the number is in base 7. Then to maximize the number,
is
and
is
. Therefore, the largest 7-10 double is 630 in base 7, or
in base 10.
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 |
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