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

(Solution)
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<cmath>a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n</cmath>
 
<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 <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.
  
 
== See also ==
 
== See also ==

Revision as of 22:40, 12 December 2017

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

\[2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}\]

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.

See also

2001 AIME I (ProblemsAnswer KeyResources)
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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