Difference between revisions of "2001 AIME I Problems/Problem 8"
(→Solution 2 (Guess and Check)) |
(→Solution 3) |
||
Line 34: | Line 34: | ||
==Solution 3== | ==Solution 3== | ||
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. | 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> | + | <cmath>98b+14b+2c=100a+10b+c</cmath> which simplifies to <cmath>2a=4b+c.</cmath> |
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}</math> in base 10. | 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}</math> in base 10. | ||
Latest revision as of 00:24, 3 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
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 The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, is and is . Therefore, the largest 7-10 double is 630 in base 7, or 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.