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
Expanding, we find that
or re-arranging,
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 |
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