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

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(→‎Solution 2 (Bash/Guess and Check))
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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>
 
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>
  
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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>.
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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>.  
 
<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>.  
Line 28: Line 28:
 
This suggests that <math>A</math> is less than <math>\frac{666}{2}=333</math>.  
 
This suggests that <math>A</math> is less than <math>\frac{666}{2}=333</math>.  
  
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Guess and check shows that <math>310<A<320</math>, and checking values in that range produces <math>\boxed{315}_{10}</math>.
+
Guess and check shows that <math>310<A<320</math>, and checking values in that range produces <math>\boxed{315}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:42, 9 May 2020

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 (Bash/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 $310<A<320$, and checking values in that range produces $\boxed{315}$.

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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