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

Revision as of 15:14, 14 January 2019

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

$2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}$ m (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.

@@ Line 5: / Line 5: @@
 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>
+<cmath>2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}</cmath>m (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

2001 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2001 AIME I Problems/Problem 8"

Revision as of 15:14, 14 January 2019

Problem

Solution

See also