Difference between revisions of "2001 AIME I Problems/Problem 8"
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Guess and check shows that <math>A<320</math>, and checking values in that range produces <math>\boxed{315}</math>. | Guess and check shows that <math>A<320</math>, and checking values in that range produces <math>\boxed{315}</math>. | ||
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+ | ==Solution 3==7, or | ||
+ | Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be <cmath>abc</cmath> in base 7. Then the number in expanded form is <cmath>49a+7b+c</cmath> in base 7 and <cmath>100a+10b+c</cmath> in base 10. Since the number in base 7 is half the number in base 10, we get the following equation. | ||
+ | <cmath>98b+14b+2c=100a+10b+c</cmath>, which simplifies to <math></math>2a=4b+c<math>. | ||
+ | The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, </math>b<math> is </math>3<math> and </math>c<math> is </math>0<math>. Therefore, the largest 7-10 double is 630 in base 7, or </math>\boxed{315}$ in base 10. | ||
== See also == | == See also == |
Revision as of 23:23, 2 August 2021
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.
Solution 2 (Guess and Check)
Let be the base representation of our number, and let be its base representation.
Given this is an AIME problem, . If we look at in base , it must be equal to , so when is looked at in base
If in base is less than , then as a number in base must be less than .
is non-existent in base , so we're gonna have to bump that down to .
This suggests that is less than .
Guess and check shows that , and checking values in that range produces .
==Solution 3==7, or Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be in base 7. Then the number in expanded form is in base 7 and in base 10. Since the number in base 7 is half the number in base 10, we get the following equation. , which simplifies to $$ (Error compiling LaTeX. ! Missing $ inserted.)2a=4b+cb3c0\boxed{315}$ in base 10.
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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