Difference between revisions of "2001 AIME I Problems/Problem 8"

Revision as of 23:23, 2 August 2021

Problem

Call a positive integer $N$ a 7-10 double if the digits of the base- $7$ representation of $N$ form a base- $10$ number that is twice $N$ . For example, $51$ is a 7-10 double because its base- $7$ representation is $102$ . What is the largest 7-10 double?

Solution

We let $N_7 = \overline{a_na_{n-1}\cdots a_0}_7$ ; we are given that

$2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}$ (This is because the digits in $N$ ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)

Expanding, we find that

$2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0$

or re-arranging,

$a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n$

Since the $a_i$ s are base- $7$ digits, it follows that $a_i < 7$ , and the LHS is less than or equal to $30$ . Hence our number can have at most $3$ digits in base- $7$ . Letting $a_2 = 6$ , we find that $630_7 = \boxed{315}_{10}$ is our largest 7-10 double.

Solution 2 (Guess and Check)

Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation.

Given this is an AIME problem, $A<1000$ . If we look at $B$ in base $10$ , it must be equal to $2A$ , so $B<2000$ when $B$ is looked at in base $10.$

If $B$ in base $10$ is less than $2000$ , then $B$ as a number in base $7$ must be less than $2*7^3=686$ .

$686$ is non-existent in base $7$ , so we're gonna have to bump that down to $666_7$ .

This suggests that $A$ is less than $\frac{666}{2}=333$ .

Guess and check shows that $A<320$ , and checking values in that range produces $\boxed{315}$ .

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 $abc$ in base 7. Then the number in expanded form is $49a+7b+c$ in base 7 and $100a+10b+c$ in base 10. Since the number in base 7 is half the number in base 10, we get the following equation. $98b+14b+2c=100a+10b+c$ , which simplifies to $2a=4b+c$ . The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, $b$ is $3$ and $c$ is $0$ . Therefore, the largest 7-10 double is 630 in base 7, or $\boxed{315}$ in base 10.

@@ Line 32: / Line 32: @@
-==Solution 3==7, or
+==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>98b+14b+2c=100a+10b+c</cmath>, which simplifies to <math></math>2a=4b+c<math>.
+<cmath>98b+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}$ in base 10.
+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 ==

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

Page

Toolbox

Search