Difference between revisions of "2001 AIME I Problems/Problem 8"
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We let <math>N_7 = \overline{a_na_{n-1}\cdots a_0}_7</math>; we are given that | 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> (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 | Expanding, we find that |
Revision as of 16:27, 7 March 2020
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
(This is because the digits in
' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)
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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