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

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== Problem ==
 
== Problem ==
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Call a positive integer <math>N</math> a <math>\textit{7-10 double}</math> if the digits of the base-7 representation of <math>N</math> form a base-10 number that is twice <math>N</math>. For example, <math>51</math> is a 7-10 double because its base-7 representation is <math>102</math>. What is the largest 7-10 double?
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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 ==
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{{solution}}
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We let <math>N_7 = \overline{a_na_{n-1}\cdots a_0}_7</math>; we are given that
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<cmath>2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}</cmath>
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Expanding, we find that
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<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>
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or re-arranging,
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<cmath>a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n</cmath>
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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 <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 ==
 
{{AIME box|year=2001|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2001|n=I|num-b=7|num-a=9}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 22:28, 11 June 2008

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 $30$. 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 (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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