Difference between revisions of "2001 AIME I Problems/Problem 8"
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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 
a 7-10 double if the digits of the base-
representation of form a base-
number that is twice . For example, 
is a 7-10 double because its base- representation is 
. What is the largest 7-10 double?
Solution
We let 
; we are given that
![\[2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}\]](http://latex.artofproblemsolving.com/8/0/e/80e28e31aaf18ab0eafe5466283a750f9e062edd.png)
Expanding, we find that
![\[2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0\]](http://latex.artofproblemsolving.com/8/3/c/83c32c1f5624df09afd480557f3483b8f0ed4190.png)
or re-arranging,
![\[a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n\]](http://latex.artofproblemsolving.com/4/5/7/4574738c53d25fa109c96f5340f1e386649da59d.png)
Since the 
s are base- digits, it follows that 
, and the LHS is less than or equal to 
. Hence our number can have at most 
digits in base-. Letting 
, we find that 
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
